ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

CHAPTER 3 KINEMATICS IN TWO DIMENSIONS

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (a) The horizontal component vx of the projectile’s velocity remains constant throughout the

motion, since the acceleration ax in the horizontal direction is zero (ax = 0 m/s2). The vertical component vy, however, changes as the projectile moves. This component is greatest at point 1, decreases to zero at point 2 at the top of the trajectory, and then increases to a magnitude less than that at point 1 as the projectile approaches point 3.

2. (b) The minimum speed of the projectile occurs when it is at the top of its trajectory. At this

point the vertical component of its velocity is zero (vy = 0 m/s). Since there is no

acceleration in the x direction (ax = 0 m/s2), the x component of the projectile’s velocity remains constant at vx = +30 m/s throughout the motion. Thus, the minimum speed is 30 m/s.

3. (c) The acceleration due to gravity is the same for both balls, despite the fact that they have different velocities.

4. (d) The acceleration of a projectile is the same at all points on the trajectory. It points downward, toward the earth, and has a magnitude of 9.80 m/s2.

5. (c) Since there is no acceleration in the x direction (ax = 0 m/s2) for projectile motion, the x component vx of the velocity is constant throughout the motion. And, the acceleration due to

gravity, ay = −9.80 m/s2 (“downward” is the negative direction), also remains constant.

6. (c) The time for a projectile to reach the ground depends only on the y component (or vertical component) of its variables, i.e., y, v0y, and ay. These variables are the same for both balls. The fact that Ball 1 is moving horizontally at the top of its trajectory does not play a role in the time it takes for it to reach the ground.

7. (a) Using y = −19.6 m, ay = −9.80 m/s2, and v0y = 0 m/s, Equation 3.5b ( )210 2y yy v t a t= +

can be used to calculate the time t.

8. (b) The time a projectile is in the air is equal to twice the time it takes to fall from its maximum height. Both have the same maximum height, so the time of fall is the same. Therefore, both projectiles are in the air for the same amount of time.

98 KINEMATICS IN TWO DIMENSIONS

9. (a) The time a projectile is in the air is equal to twice the time it takes to fall from its maximum height. Projectile 1 reaches the greater height, so it spends the greater amount of time in the air.

10. (c) The vertical component (or y component) of the final velocity depends on the y components of the kinematic variables ( y, v0y, and ay) and the time t. These variables are the same for both balls, so they have the same vertical component of the velocity.

11. (a) A person standing on the ground sees a car traveling at +25 m/s. When the driver of the car looks out the window, she sees the person moving in the opposite direction with the same speed, or with a velocity of −25 m/s.

12. (d) The velocity vBC of the bus relative to the car is given by the relation vBC = vBG + vGC = vBG + (−vCG) = +16 m/s + (−12 m/s) = +4 m/s.

13. (d) This answer is arrived at by using the relation vPG = vPB + vBG = +2 m/s + 16 m/s = +18 m/s.

14. (c) The velocity vPC of the passenger relative to the car is given by vPC = vPB + vBC, according to the subscripting method discussed in Section 3.4. However, the last term on the right of this equation is given by vBC = vBG + vGC. So, vPC = vPB + vBG + vGC = +2 m/s + 16 m/s + (−12 m/s) = +6 m/s.

15. (b) The velocity of the jeep relative to you is zero. Thus, the horizontal component of the tire’s velocity relative to you is also zero. Since this component of the velocity never changes as the tire falls, the car cannot hit the tire, regardless of how close the car is to the jeep.

16. The magnitude vAB of the velocity of car A relative to car B is vAB = 34.2 m/s. The angle

that the velocity vAB makes with respect to due east is θ = 37.9° south of east.

Chapter 3 Problems 99

CHAPTER 3 KINEMATICS IN TWO DIMENSIONS

PROBLEMS

1. SSM REASONING The displacement is a vector drawn from the initial position to the

final position. The magnitude of the displacement is the shortest distance between the positions. Note that it is only the initial and final positions that determine the

displacement. The fact that the squirrel jumps to an intermediate position before reaching his final position is not important. The trees are perfectly straight and both growing perpendicular to the flat horizontal ground beneath them. Thus, the distance between the trees and the length of the trunk of the second tree below the squirrel’s final landing spot form the two perpendicular sides of a right triangle, as the drawing shows. To this triangle, we can apply the Pythagorean theorem and determine the magnitude A of the displacement vector A.

SOLUTION According to the Pythagorean theorem, we have

( ) ( )2 21.3 m 2.5 m 2.8 mA= + =

2. REASONING The meteoroid’s speed is the magnitude of its velocity vector, here described

in terms of two perpendicular components, one directed toward the east and one directed vertically downward. Let east be the +x direction, and up be the +y direction. Then the components of the meteoroid’s velocity are vx = +18.3 km/s and vy = −11.5 km/s. The meteoroid’s speed v is related to these components by the Pythagorean theorem (Equation 1.7): 2 2 2

x yv v v= + . SOLUTION From the Pythagorean theorem,

( ) ( )2 22 2 18.3 km/s 11.5 km/s 21.6 km/sx yv v v= + = + + − =

It’s important to note that the negative sign for vy becomes a positive sign when this quantity is squared. Forgetting this fact would yield a value for v that is smaller than vx, but the magnitude of a vector cannot be smaller than either of its components.

A

1.3 m

2.5 m


3. REASONING To determine the horizontal and vertical components of the launch velocity,

we will use trigonometry. To do so, however, we need to know both the launch angle and the magnitude of the launch velocity. The launch angle is given. The magnitude of the launch velocity can be determined from the given acceleration and the definition of acceleration given in Equation 3.2.

SOLUTION According to Equation 3.2, we have

a = v − v0

t − t0 or 340 m/s2 = v − 0 m/s

0.050 s or v = 340 m/s2( ) 0.050 s( )

Using trigonometry, we find the components to be

vx = v cos 51° = 340 m/s2( ) 0.050 s( )cos 51° = 11 m/s

vy = v sin 51° = 340 m/s2( ) 0.050 s( )sin 51° = 13 m/s

4. REASONING The displacement is the vector drawn from the initial to the final position.

The magnitude of the displacement vector is the shortest distance between the initial and final position.

SOLUTION The batter’s initial position is home plate. His final position is at third base. The shortest distance between these two positions is 27.4 m. Therefore, the magnitude of the player’s displacement is 27.4 m .

____________________________________________________________________________________________ 5. SSM REASONING The displacement of the elephant seal

has two components; 460 m due east and 750 m downward. These components are mutually perpendicular; hence, the Pythagorean theorem can be used to determine their resultant.

SOLUTION From the Pythagorean theorem,

2 2 2(460 m) (750 m)R = +

Therefore,

R = + = ×( (460 m) 750 m) 8.82 2 102 m 6. REASONING AND SOLUTION The horizontal displacement is


x = 19 600 m – 11 200 m = 8400 m

The vertical displacement is

y = 4900 m – 3200 m = 1700 m The magnitude of the displacement is therefore,

( ) ( )2 22 2 8400 m 1700 m 8600 mr x yΔ = + = + = ____________________________________________________________________________________________ 7. SSM REASONING AND SOLUTION

x = r cos θ = (162 km) cos 62.3° = 75.3 km

y = r sin θ = (162 km) sin 62.3° = 143 km ____________________________________________________________________________________________ 8. REASONING Consider first the shopper’s ride up the escalator. Let the diagonal length of

the escalator be L, the height of the upper floor be H, and the angle that the escalator makes with respect to the horizontal be θ (see the diagram). Because L is the hypotenuse of the right triangle and H is opposite the angle θ, the three quantities are related by the inverse sine function:

1 1osin sinh Hh L

θ − −⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ (1.4)

Now consider the entire trip from the bottom to the top of the escalator (a distance L), and then from the top of the escalator to the store entrance (a distance s). The right turn between these two parts of the trip means that they are perpendicular (see the diagram). The shopper’s total displacement has a magnitude D, and this serves as the hypotenuse of a right triangle with L and s. From the Pythagorean theorem, the three sides are related as follows: 2 2 2D L s= + . SOLUTION Solving 2 2 2D L s= + for the length L of the escalator gives 2 2L D s= − . We

now use this result and the relation 1sin HL

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠ to obtain the angle θ:

( ) ( )1 1 1

2 2 2 2

6.00 msin sin sin 27.016.0 m 9.00 m

H HL D s

θ − − −⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟= = = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎜ ⎟−⎝ ⎠ −⎝ ⎠

o

θ

D

L H L

s

Entire view

Up the escalator


9. SSM REASONING a. We designate the direction down and parallel to the ramp as the +x direction, and the

table shows the variables that are known. Since three of the five kinematic variables have values, one of the equations of kinematics can be employed to find the acceleration ax.

x-Direction Data x ax vx v0x t

+12.0 m ? +7.70 m/s 0 m/s b. The acceleration vector points down and parallel to the ramp, and the

angle of the ramp is 25.0° relative to the ground (see the drawing). Therefore, trigonometry can be used to determine the component aparallel of the acceleration that is parallel to the ground.

SOLUTION a. Equation 3.6a ( )2 2

0 2x x xv v a x= + can be used to find the acceleration in terms of the three known variables. Solving this equation for ax gives

( ) ( )

( )

2 22 220 7.70 m/s 0 m/s

2.47 m/s2 2 12.0 m

x xx

v va

x+ −−

= = =+

b. The drawing shows that the acceleration vector is oriented 25.0° relative to the ground. The component aparallel of the acceleration that is parallel to the ground is

( )2 2

parallel = cos 25.0 2.47 m/s cos 25.0 2.24 m/sxa a ° = ° = 10. REASONING The component method can be used to determine the magnitude and

direction of the bird watcher's displacement. Once the displacement is known, Equation 3.1 can be used to find the average velocity.

SOLUTION The following table gives the components of the individual displacements of

the bird watcher. The last entry gives the components of the bird watcher's resultant displacement. Due east and due north have been chosen as the positive directions.

25.0° aparallel

ax


Displacement

East/West Component

North/South Component

A 0.50 km 0 B C

0 –(2.15 km) cos 35.0° = –1.76 km

–0.75 km (2.15 km) sin 35.0° = 1.23 km

r = A + B + C

–1.26 km

0.48 km

a. From the Pythagorean theorem, we have

r = + =(–1.26 km) (0.48 km) 1.35 km2 2

The angle θ is given by

θ = tan−1 0.48 km1.26 km

⎛⎝⎜

⎞⎠⎟ = 21°, north of west

b. From Equation 3.1, the average velocity is

1.35 km 0.540 km/h, 21 north of west2.50 ht

Δ= = = °Δrv

Note that the direction of the average velocity is, by definition, the same as the direction of

the displacement. 11. REASONING AND SOLUTION

a. Average speed is defined as the total distance d covered divided by the time Δt required to cover the distance. The total distance covered by the earth is one-fourth the circumference of its circular orbit around the sun:

d =

14 x 2π (1.50 x 1011 m) = 2.36 x 1011 m

v dt

= = ××

= ×Δ

2 36 1011. m 7.89 10

2.99 10 m / s64

s

b. The average velocity is defined as the displacement

divided by the elapsed time.


In moving one-fourth of the distance around the sun, the earth completes the displacement shown in the figure at the right. From the Pythagorean theorem, the magnitude of this displacement is

Δr = r2 + r2 = 2 r Thus, the magnitude of the average velocity is

r

r

rΔ

v r= = × ××

= ×ΔΔt

2 1 50 1011. m 7.89 10

2.69 10 m / s64

s

12. REASONING The motion in the x direction occurs independently of the motion in the y direction. The components of the velocity change from their initial values of v0x and v0y to their final values of vx and vy. The changes occur, respectively, because of the acceleration components ax and ay. The final values can be determined with the aid of Equations 3.3a and 3.3b.

SOLUTION a. According to Equation 3.3a, the x component of the velocity is

( )( )2

0 5480 m/s + 1.20 m/s 842 s 6490 m/sx x xv v a t= + = = b. According to Equation 3.3b, the y component of the velocity is

( )( )20 0 m/s + 8.40 m/s 842 s 7070 m/sy y yv v a t= + = =

____________________________________________________________________________________________ 13. SSM REASONING The vertical component of the ball’s velocity v0 changes as the ball

approaches the opposing player. It changes due to the acceleration of gravity. However, the horizontal component does not change, assuming that air resistance can be neglected. Hence, the horizontal component of the ball’s velocity when the opposing player fields the ball is the same as it was initially.

SOLUTION Using trigonometry, we find that the horizontal component is

vx = v0 cosθ = 15 m/s( )cos 55° = 8.6 m/s

14. REASONING We will determine the ball’s displacement y in the vertical direction from 1 2

0 2y yy v t a t= + (Equation 3.5b). We choose this equation because we know that


0 0.00 m / syv = (the ball initially travels horizontally) and that 29.80 m/sya = − (assuming that upward is the +y direction). We do not know the value of the ball’s time of flight t. However, we can determine it from the motion in the horizontal or x direction.

SOLUTION From Equation 3.5b we have that

( )1 1 12 2 20 2 2 2

0.00 m / sy y y yy v t a t t a t a t= + = + =

The motion in the horizontal direction occurs at a constant velocity of 0 28.0 m / sxv = + and the displacement in the horizontal direction is 19.6 mx = + . Thus, it follows that

00

19.6 m or 0.700 s28.0 m / sx

x

xx v t tv

= = = =

Using this value for the time in the expression for y shows that

( )( )21 12 22 2

9.80 m/s 0.700 s 2.40 myy a t= = − = −

Since the displacement of the ball is 2.40 m downward, the ball is 2.40 m above the court when it leaves the racket.

_____________________________________________________________________________________________ 15. REASONING

a. Here is a summary of the data for the skateboarder, using v0 = 6.6 m/s and θ = 58°:

y-Direction Data y ay vy v0y t

? −9.80 m/s2 0 m/s +(6.6 m/s)sin 58° = +5.6 m/s


? 0 m/s2 +(6.6 m/s)cos 58° = +3.5 m/s

We will use the relation 2 20 2y y yv v a y= + (Equation 3.6b) to find the skateboarder’s vertical

displacement y above the end of the ramp. When the skateboarder is at the highest point, his vertical displacement y1 above the ground is equal to his initial height y0 plus his vertical displacement y above the end of the ramp: y1 = y0 + y. Next we will use 0y y yv v a t= +


(Equation 3.3b) to calculate the time t from the launch to the highest point, which, with 21

0 2x xx v t a t= + (Equation 3.5a), will give us his horizontal displacement x.

SOLUTION a. Substituting vy = 0 m/s into Equation 3.6b and solving for y, we find

( ) ( )( )

2 22 02

0 25.6 m/s

0 m/s 2 y or 1.6 m2 2 9.80 m/s

yy y

y

vv a y

a− −

= + = = = +−

Therefore, the highest point y1 reached by the skateboarder occurs when y1 = y0 + y = 1.2 m + 1.6 m = +2.8 m above the ground.

b. We now turn to the horizontal displacement, which requires that we first find the elapsed time t. Putting vy = 0 m/s into Equation 3.3b and solving for t yields

0

00 m/s or yy y

y

vv a t t

a

−= + = (1)

Given that ax = 0 m/s2, the relation 21

0 2x xx v t a t= + (Equation 3.5a) reduces to 0xx v t= . Substituting Equation (1) for t then gives the skateboarder’s horizontal displacement:

( )( )0 0 00 0

3.5 m/s 5.6 m/s2.0 m

9.80 m/sy x y

x xy y

v v vx v t v

a a

⎛ ⎞− − − + += = = = = +⎜ ⎟⎜ ⎟ −⎝ ⎠

16. REASONING The magnitude v of the puck’s velocity is related to its x and y velocity

components (vx and vy ) by the Pythagorean theorem (Equation 1.7); 2 2x yv v v= + . The

relation 0x x xv v a t= + (Equation 3.3a) may be used to find vx, since v0x, ax, and t are known. Likewise, the relation 0y y yv v a t= + (Equation 3.3b) may be employed to determine vy, since v0y, ay, and t are known. Once vx and vy are determined, the angle θ that the velocity makes with respect to the x axis can be found by using the inverse tangent function (Equation 1.6); ( )1tan /y xv vθ −= .


SOLUTION Using Equations 3.3a and 3.3b, we find that

( )( )

( )( )

20

20

1.0 m/s 2.0 m/s 0.50 s 2.0 m/s

2.0 m/s 2.0 m/s 0.50 s 1.0 m/s

x x x

y y y

v v a t

v v a t

= + = + + = +

= + = + + − = +

The magnitude v of the puck’s velocity is

( ) ( )2 22 2 2.0 m/s 1.0 m/s 2.2 m/sx yv v v= + = + =

The angle θ that the velocity makes with respect to the +x axis is

1 1 1.0 m/stan tan 27 above the + axis2.0 m/s

y

x

vx

vθ − −⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

_____________________________________________________________________________________________ 17. SSM REASONING Since the spider encounters no appreciable air resistance during its

leap, it can be treated as a projectile. The thickness of the magazine is equal to the spider’s vertical displacement y during the leap. The relevant data are as follows (assuming upward to be the +y direction):

y-Direction Data

y ay vy v0y t

? −9.80 m/s2 (0.870 m/s) sin 35.0° = +0.499 m/s 0.0770 s

SOLUTION We will calculate the spider’s vertical displacement directly from 21

0 2y yy v t a t= + (Equation 3.5b):

( )( ) ( )( )22120.499 m/s 0.0770 s 9.80 m/s 0.0770 s 0.0094 my = + − =

To express this result in millimeters, we use the fact that 1 m equals 1000 mm:

Thickness 0.0094 m= ( ) 1000 mm1 m9.4 mm

⎛ ⎞=⎜ ⎟

⎝ ⎠

18. REASONING The motion of the bullet in the horizontal direction occurs at a constant velocity 0 670 m / sxv = , so that we can use 0xx v t= to determine the displacement x between the end of the rifle and the bull’s-eye. The magnitude of x is the distance we seek. However, we also need to know the time of flight t in order to determine x. We will


determine t from the bullet’s displacement y in the vertical direction by using 1 2

0 2y yy v t a t= + (Equation 3.5b). We choose this equation because we know that

0.025 my = − (assuming that upward is the +y direction), 0 0.00 m / syv = (the bullet

initially travels horizontally) and 29.80 m/sya = − . Thus, we know every variable in the equation except t.

SOLUTION The displacement x between the end of the rifle and the bull’s-eye is

0xx v t= Referring to Equation 3.5b for the vertical motion of the bullet and the fact

that 0 0.00 m / syv = , we have

( )1 1 12 2 20 2 2 2

20.00 m / s or y y y yy

yy v t a t t a t a t ta

= + = + = =

Substituting this result into the equation for x gives

( ) ( )0 0 2

2 0.025 m2 670 m / s 48 m9.80 m/sx x

y

yx v t va

−= = = =

−

Thus, the distance between the end of the rifle and the bull’s-eye is 48 m ____________________________________________________________________________________________ 19. REASONING a. The maximum possible distance that the ball can travel occurs when it is launched at an

angle of 45.0°. When the ball lands on the green, it is at the same elevation as the tee, so the vertical component (or y component) of the ball's displacement is zero. The time of flight is given by the y variables, which are listed in the table below. We designate "up" as the +y direction.

y-Direction Data y ay vy v0y t

0 m −9.80 m/s2 +(30.3 m/s) sin 45.0° = +21.4 m/s ?

Since three of the five kinematic variables are known, we can employ one of the equations of kinematics to find the time t that the ball is in the air.

b. The longest hole in one that the golfer can make is equal to the range R of the ball. This distance is given by the x variables and the time of flight, as determined in part (a). Once again, three variables are known, so an equation of kinematics can be used to find the range of the ball. The +x direction is taken to be from the tee to the green.


x-Direction Data

x ax vx v0x t

R = ? 0 m/s2 +(30.3 m/s) cos 45.0° = +21.4 m/s from part a

SOLUTION a. We will use Equation 3.5b to find the time, since this equation involves the three known variables in the y direction:

( )( )

21 12 20 0

212

=

0 m = +21.4 m/s + 9.80 m/s

y y y yy v t a t v a t t

t t

+ = +

⎡ ⎤−⎣ ⎦

Solving this quadratic equation yields two solutions, t = 0 s and t = 4.37 s. The first solution represents the situation when the golf ball just begins its flight, so we discard this one. Therefore, 4.37 st = .

b. With the knowledge that t = 4.37 s and the values for ax and v0x (see the x-direction data table above), we can use Equation 3.5a to obtain the range R of the golf ball.

x =R = v

0xt + 1

2 axt2 = +21.4 m/s( ) 4.37 s( ) + 1

2 0 m/s2( ) 4.37 s( )2= 93.5 m

______________________________________________________________________________ 20. REASONING The magnitude (or speed) v of the ball’s velocity is related to its x and y

velocity components (vx and vy ) by the Pythagorean theorem; 2 2x yv v v= +

(Equation 1.7).

The horizontal component vx of the ball’s velocity never changes during the flight, since, in the absence of air resistance, there is no acceleration in the x direction (ax = 0 m/s2). Thus, vx

is equal to the horizontal component v0x of the initial velocity, or 0 0 cos40.0x xv v v= = o. Since v0 is known, vx can be determined.

The vertical component vy of the ball’s velocity does change during the flight, since the

acceleration in the y direction is that due to gravity (ay = −9.80 m/s2). The relation 2 2

0 2y y yv v a y= + (Equation 2.6b) may be used to find 2yv , since ay, y, and v0y are known

(v0y = v0 sin 40.0º).


SOLUTION The speed v of the golf ball just before it lands is

v = vx2 + vy

2 = v0 cos40.0°( )2 + v0 y2 + 2ay y

vy

2

= v0 cos40.0°( )2 + v0 sin40.0°( )2 + 2ay y

= 14.0 m/s( )cos40.0°⎡⎣ ⎤⎦2+ 14.0 m/s( )sin40.0°⎡⎣ ⎤⎦

2+ 2 −9.80 m/s2( ) 3.00 m( ) = 11.7 m/s

21. SSM REASONING When the skier leaves the ramp, she exhibits projectile motion. Since

we know the maximum height attained by the skier, we can find her launch speed v0 using Equation 3.6b , 2 2

0 2y y yv v a y= + , where 0 0 sin 63yv v= °. SOLUTION At the highest point in her trajectory, vy = 0. Solving Equation 3.6b for v0y we

obtain, taking upward as the positive direction,

v v a y va y

y yy

0 0 06380 13

= °= =°=

°=sin –2

–2

sin–2(–9. )

sin or

63 m / s ( m)

6318 m / s

2

____________________________________________________________________________________________


22. REASONING The vehicle’s initial velocity is in the +y direction, and it accelerates only in the +x direction. Therefore, the y component of its velocity remains constant at vy = +21.0 m/s. Initially, the x component of the vehicle’s velocity is zero (v0x = 0 m/s), but it increases at a rate of ax = +0.320 m/s2, reaching a final value vx given by the relation

0x x xv v a t= + (Equation 3.3a) when the pilot shuts off the RCS thruster. Once we have found vx, we will use the right triangle shown in the drawing to find the magnitude v and direction θ of the vehicle’s final velocity.

SOLUTION a. During the 45.0-second thruster burn, the x component of the vehicle’s velocity increases from zero to

( ) ( )( )20 0 m/s 0.320 m/s 45.0 s 14.4 m/sx x xv v a t= + = + =

Applying the Pythagorean theorem to the right triangle in the drawing, we find the magnitude of the vehicle’s velocity to be:

( ) ( )2 22 2 14.4 m/s 21.0 m/s 25.5 m/sx yv v v= + = + =

b. Referring again to the drawing, we see that the x and y components of the vehicle’s velocity are related by the tangent of the angle θ. Therefore, the angle of the vehicle’s final velocity is

1 1 14.4 m/stan tan 34.421.0 m/s

x

y

vv

θ − −⎛ ⎞ ⎛ ⎞⎜ ⎟= = =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

o

23. SSM REASONING Since the magnitude of the velocity of the fuel tank is given by

v v vx y= +2 2 , it is necessary to know the velocity components xv and yv just before impact. At the instant of release, the empty fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are given by 0 0 cosxv v θ= and 0 0 sinyv v θ= , where 0v is the speed of the plane at the instant of release. Since the x motion has zero acceleration, the x component of the velocity of the plane remains equal to 0xv for all later times while the tank is airborne. The y component of the velocity of the tank after it has undergone a vertical displacement y is given by Equation 3.6b.

v vy

vx

θ

+x +y


SOLUTION a. Taking up as the positive direction, the velocity components of the fuel tank just before it

hits the ground are

20 cos (135 m/s) cos 15 1.30 10 m/sx xv v v θ= = = ° = ×

From Equation 3.6b, we have

v v a y v a yy y y y= + = +

= ° + − − × =

– – ( sin )

– ( ( . )( –201

02

02

2

2 2

135 2 9 80 10

θ

m / s) sin 15.0 m / s 2.00 m / s2 3 m)

Therefore, the magnitude of the velocity of the fuel tank just before impact is

v v vx y= + = × + =2 2 10 201(1.30 m / s) m / s) 239 m / s2 22 (

The velocity vector just before impact is inclined at an angle φ with respect to the horizontal. This angle is

φ = tan−1 201 m/s

1.30×102 m/s⎛⎝⎜

⎞⎠⎟= 57.1°

b. As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the

same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the velocity of the fuel tank in part b just before impact is 239 m/s at an angle of 57.1 with respect to the horizontal° .

____________________________________________________________________________________________ 24. REASONING In the absence of air resistance, the horizontal velocity component never

changes from its initial value of v0x. Therefore the horizontal distance D traveled by the criminal (which must equal or exceed the distance between the two buildings) is the initial velocity times the travel time t, or 0xD v t= .

The time t can be found by noting that the motion of the criminal between the buildings is that of a projectile whose acceleration in the y direction is that due to gravity (ay = −9.80 m/s2, assuming downward to be the negative direction). The relation

210 2y yy v t a t= + (Equation 3.5b) allows us to determine the time, since y, v0y, and ay are

known. Since the criminal is initially running in the horizontal direction, v0y = 0 m/s. Setting

v0y = 0 m/s and solving the equation above for t yields 2 / yt y a= . In this result,

y = −2.0 m, since downward is the negative direction.


SOLUTION The horizontal distance traveled after launch is 0xD v t= . Substituting

2 / yt y a= into this relation gives

( ) ( )0 0 2

2 m2 5.3 m/s 3.4 m m/sx x

y

yD v t va

−2.0= = = =

−9.8

25. REASONING The data given in the problem are summarized as follows:

With these data, we can use Equations 3.5a and 3.5b to determine the acceleration

component ax and ay. SOLUTION Using Equations 3.5a and 3.5b, we can solve for the acceleration components

and find that

( ) ( )( )( )

1 2 00 22

6

2

2

2 2 or (3.5a)

2 4.11 10 m 2 4370 m/s 684 s

684 s

4.79 m/s

xx x x

x

x v tx v t a t a

t

a

−= + =

× −=

=

( ) ( )( )( )

01 20 22

6

2

2

2 2 or (3.5b)

2 6.07 10 m 2 6280 m/s 684 s

684 s

7.59 m/s

yy y y

y

y v ty v t a t a

t

a

−= + =

× −=

=

26. REASONING In the absence of air resistance, there is no acceleration in the x direction

(ax = 0 m/s2), so the range R of a projectile is given by R = v0x t, where v0x is the horizontal component of the launch velocity and t is the time of flight. We will show that v0x and t are each proportional to the initial speed v0 of the projectile, so the range is proportional to 2

0v .

y-Direction Data

y ay vy v0y t

6.07 × 106 m ? 6280 m/s 684 s

x-Direction Data

x ax vx v0x t

4.11 × 106 m ? 4370 m/s 684 s


Since v0x = v0 cos θ, where θ is the launch angle, we see that v0x is proportional to v0. To show that the time of flight t is also proportional to the launch speed v0, we use the fact that, for a projectile that is launched from and returns to ground level, the vertical displacement is y = 0 m. Using the relation 21

0 2y yy v t a t= + (Equation 3.5b), we have

021 10 02 2

20 m or 0 m or y

y y y yy

vv t a t v a t t

a−

= + = + =

Thus, the flight time is proportional to the vertical component of the launch velocity v0y, which, in turn, is proportional to the launch speed v0 since v0y = v0 sin θ.

SOLUTION The range of a projectile is given by R = v0x t and, since both v0x and t are

proportional to v0, the range is proportional to 20v . The given range is 23 m. When the

launch speed doubles, the range increases by a factor of 22 = 4, since the range is proportional to the square of the speed. Thus, the new range is

R = 4 23 m( ) = 92 m 27. SSM REASONING AND SOLUTION The water exhibits projectile motion. The x

component of the motion has zero acceleration while the y component is subject to the acceleration due to gravity. In order to reach the highest possible fire, the displacement of the hose from the building is x, where, according to Equation 3.5a (with ax = 0 m/s2),

0 0( cos )xx v t v tθ= =

with t equal to the time required for the water the reach its maximum vertical displacement.

The time t can be found by considering the vertical motion. From Equation 3.3b,

0y y yv v a t= + When the water has reached its maximum vertical displacement, vy = 0 m/s. Taking up and

to the right as the positive directions, we find that

0 0 siny

y y

v vt

a aθ− −

= =

and

00

sin( cos )

y

vx v

aθ

θ⎛ ⎞−⎜ ⎟=⎜ ⎟⎝ ⎠


Therefore, we have

2 20

2cos sin (25.0 m/s) cos 35.0 sin 35.0 30.0 m

–9.80 m/sy

vx

aθ θ ° °= − = − =

28. REASONING We will treat the horizontal and vertical parts

of the motion separately. The directions upward and to the right are chosen as the positive directions in the drawing. Ignoring air resistance, we note that there is no acceleration in the horizontal direction. Thus, the horizontal component v0x of the ball’s initial velocity remains unchanged

throughout the motion, and the horizontal component x of the displacement is simply v0x times the time t during which the motion occurs. This is true for both bunted balls. Since the value of x is the same for both, we have that

AA B 0 ,A A 0 ,B B 0 ,B 0 ,A

B or or x x x x

tx x v t v t v v

t⎛ ⎞

= = = ⎜ ⎟⎜ ⎟⎝ ⎠ (1)

To use this result to calculate v0x,B, it is necessary to determine the times tA and tB. We can

accomplish this by applying the equations of kinematics to the vertical part of the motion for each ball. The data for the vertical motion are summarized as follows:

Note that the initial velocity components v0y are zero, because the balls are bunted

horizontally. With these data, Equation 3.5b gives

( )1 12 20 2 2

20 m/s or y y yy

yy v t a t t a t ta

= + = + = (2)

SOLUTION Using Equation (2) for each ball and substituting the expressions for tA and tB

into Equation (1) gives

( )

A ,AA0 ,B 0 ,A 0 ,A

B B ,B

A0 ,A

B

2 /

2 /

1.2 m1.9 m/s 1.7 m/s1.5 m

yx x x

y

x

y atv v v

t y a

yv

y

⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠

= = =

y-Direction Data, Ball B

yB ay,B vy,B v0y,B tB

−1.5 m −9.80 m/s2 0 m/s ?

y-Direction Data, Ball A

yA ay,A vy,A v0y,A tA

−1.2 m −9.80 m/s2 0 m/s ?

v0x

x

y


Note that the accelerations ay,A and ay,B both equal the acceleration due to gravity and are

eliminated algebraically from the calculation. 29. REASONING The vertical displacement y of the ball depends on the time that it is in the air

before being caught. These variables depend on the y-direction data, as indicated in the table, where the +y direction is "up."

y-Direction Data

y ay vy v0y t

? −9.80 m/s2 0 m/s ?

Since only two variables in the y direction are known, we cannot determine y at this point. Therefore, we examine the data in the x direction, where +x is taken to be the direction from the pitcher to the catcher.

Since this table contains three known variables, the time t can be evaluated by using an equation of kinematics. Once the time is known, it can then be used with the y-direction data, along with the appropriate equation of kinematics, to find the vertical displacement y.

SOLUTION Using the x-direction data, Equation 3.5a can be employed to find the time t

that the baseball is in the air:

( )2 2120 0 since 0 m/sx x x xx v t a t v t a= + = =

Solving for t gives

0

17.0 m 0.415 s+41.0 m/sx

xtv

+= = =

The displacement in the y direction can now be evaluated by using the y-direction data table

and the value of t = 0.415 s. Using Equation 3.5b, we have

( )( ) ( )( )22 21 12 20 0 m/s 0.415 s 9.80 m/s 0.415 s 0.844 my yy v t a t= + = + − = −

The distance that the ball drops is given by the magnitude of this result, so

Distance = 0.844 m . ______________________________________________________________________________


+17.0 m 0 m/s2 +41.0 m/s ?


30. REASONING The data for the problem are summarized below. In the tables, we use the

symbol v0 to denote the speed with which the ball is thrown and choose upward and to the right as the positive directions.

Note that ax = 0 m/s2, because air resistance is being ignored. Also note that y = 0 m,

because the football rises and then returns to the same vertical level from which it was launched. Finally, we have used trigonometry to express the components v0x and v0y of the initial velocity in terms of the speed v0 and the 30.0° launch angle. The key here is to remember that the horizontal and vertical parts of the motion can be treated separately, the time for the motion being the same for each. Since air resistance is being ignored, we can apply the equations of kinematics separately to the motions in the x and y directions.

SOLUTION Since there is no acceleration in the horizontal direction, motion in that

direction is constant-velocity motion, and the horizontal displacement x is simply the initial velocity component v0x times the time t:

( )0 0 cos30.0xx v t v t= = °

An expression for t can be obtained by considering the motion in the vertical direction.

Thus, we use Equation 3.5b from the equations of kinematics and recognize that the displacement y is zero and v0y = v0 sin 30.0°:

( )1 12 2 00 02 2

2 sin30.0 or 0 m sin30.0 or y y y

y

vy v t a t v t a t t

a− °

= + = ° + =

Substituting this result for the time into the expression for x gives

x-Direction Data

x ax vx v0x t

183 m 0 m/s2 v0 cos 30.0° ?

y-Direction Data

y ay vy v0y t

0 m −9.80 m/s2 v0 sin 30.0° Same as for x direction

30.0°

v0

183 m

v0 v0y

v0x

30.0°


( ) ( )

( )( )

20 0

0 0

2

0

2 sin 30.0 2 cos30.0 sin 30.0cos30.0 cos30.0

183 m 9.80 m/s45.5 m/s

2cos30.0 sin 30.0 2cos30.0 sin 30.0

y y

y

v vx v t v

a a

xav

⎛ ⎞− ° ° °⎜ ⎟= ° = ° = −⎜ ⎟⎝ ⎠

− −−= = =

° ° ° °

31. SSM REASONING The speed of the fish at any time t is given by v v vx y= +2 2 , where

vx and vy are the x and y components of the velocity at that instant. Since the horizontal motion of the fish has zero acceleration, 0x xv v= for all times t. Since the fish is dropped by the eagle, v0x is equal to the horizontal speed of the eagle and 0 0yv = . The y component of

the velocity of the fish for any time t is given by Equation 3.3b with 0 0yv = . Thus, the

speed at any time t is given by v v a tx y= +02 2( ) .

SOLUTION a. The initial speed of the fish is v v v v vx y x x0 0

202

02 2

00= + = + = . When the fish's speed doubles, 02 xv v= . Therefore,

2 40 0

2 202

02 2v v a t v v a tx x y x x y= + = +( ) ( )or

Assuming that downward is positive and solving for t, we have

t = 3 v0x

ay= 3 6.0 m/s

9.80 m/s2⎛⎝⎜

⎞⎠⎟ = 1.1 s

b. When the fish's speed doubles again, 04 xv v= . Therefore,

4 160 02 2

02

02 2v v a t v v a tx x y x x y= + = +( ) ( )or

Solving for t, we have

t = 15 v0x

ay= 15 6.0 m/s

9.80 m/s2⎛⎝⎜

⎞⎠⎟ = 2.37 s

Therefore, the additional time for the speed to double again is ( .2 4 s) – (1.1 s) = 1.3 s . ____________________________________________________________________________________________


32. REASONING Since we know the launch angle θ = 15.0°, the launch speed v0 can be obtained using trigonometry, which gives the y component of the launch velocity as v0y = v0 sin θ. Solving this equation for v0 requires a value for v0y, which we can obtain from the vertical height of y = 13.5 m by using Equation 3.6b from the equations of kinematics.

SOLUTION From Equation 3.6b we have

vy2 = v0y

2 + 2ayy

0 m/s( )2 = v0y2 + 2 −9.80 m/s2( ) 13.5 m( ) or v0y = 2 9.80 m/s2( ) 13.5 m( )

Using trigonometry, we find

v0 =v0y

sin 15.0°=

2 9.80 m/s2( ) 13.5 m( )sin 15.0°

= 62.8 m/s

____________________________________________________________________________________________ 33. REASONING The drawing at the right shows the

velocity vector v of the water at a point below the top of the falls. The components of the velocity are also shown. The angle θ is given by tan /y xv vθ = , so that

– tan –(2.7 m/s) tan 75 y xv v θ= = °

v

θvx

vy

xy

θ

Here we have used the fact that the horizontal velocity component vx remains unchanged at its initial value of 2.7 m/s as the water falls. Knowing the y component of the velocity, we can use Equation 3.6b, ( 2 2

0 2y y yv v a y= + ) to find the vertical distance y. SOLUTION Taking 0 0yv = m/s, and taking upward as the positive direction, we have from

Equation 3.5b that

yv

ay

y

= =°

=2

22 7 75

2–( . tan

)–5.

m / s)2(–9.80 m / s

m2

2

Therefore, the velocity vector of the water points downward at a 75° angle below the

horizontal at a vertical distance of 5.2 m below the edge. ____________________________________________________________________________________________


34. REASONING We begin by considering the flight time of the ball on the distant planet. Once the flight time is known, we can determine the maximum height and the range of the ball.

The range of a projectile is proportional to the time that the projectile is in the air.

Therefore, the flight time on the distant plant 3.5 times larger than on earth. The flight time can be found from Equation 3.3b ( 0y y yv v a t= + ). When the ball lands, it is at the same

level as the tee; therefore, from the symmetry of the motion 0–y yv v= . Taking upward and to the right as the positive directions, we find that the flight time on earth would be

0 0 02

2 2 sin 2(45 m/s) sin 29 4.45 s–9.80 m/s

y y y

y y y

v v v vt

a a aθ− − − − °= = = = =

Therefore, the flight time on the distant planet is 3.5 × (4.45 s) = 15.6 s. From the symmetry

of the problem, we know that this is twice the amount of time required for the ball to reach its maximum height, which, consequently, is 7.80 s.

SOLUTION a. The height y of the ball at any instant is given by Equation 3.4b as the product of the

average velocity component in the y direction 12 v0y + vy( ) and the time t: y = 1

2 v0y + vy( )t . Since the maximum height H is reached when the final velocity component in the y direction is zero (vy = 0 m/s), we find that

H v t v ty= = ° = °1

2 012 0

12(sin 29 ) (45 m / s) (sin 29 (7.80 s) = 85) m

b. The range of the ball on the distant planet is

x v t v tx= = ° = °0 0 (cos ( 29 ) 45 m / s) (cos 29 ) (15.6 s) = 610 m 35. REASONING The rocket will clear the top of the wall by an amount that is the height of

the rocket as it passes over the wall minus the height of the wall. To find the height of the rocket as it passes over the wall, we separate the rocket’s projectile motion into its horizontal and vertical parts and treat each one separately. From the horizontal part we will obtain the time of flight until the rocket reaches the location of the wall. Then, we will use this time along with the acceleration due to gravity in the equations of kinematics to determine the height of the rocket as it passes over the wall.

SOLUTION We begin by finding the horizontal and vertical components of the launch

velocity


v0x = v0 cos 60.0° = 75.0 m/s( )cos 60.0°

v0y = v0 sin 60.0° = 75.0 m/s( )sin 60.0°

Using v0x, we can obtain the time of flight, since the distance to the wall is known to be

27.0 m:

t = 27.0 mv0x

= 27.0 m75.0 m/s( )cos 60.0°

= 0.720 s The height of the rocket as it clears the wall can be obtained from Equation 3.5b, in which

we take upward to be the positive direction. The amount by which the rocket clears the wall can then be obtained:

y = v0yt + 12 ayt

2

y = 75.0 m/s( ) sin 60.0°( ) 0.720 s( ) + 12 −9.80 m/s2( ) 0.720 s( )2 = 44.2 m

clearance = 44.2 m −11.0 m= 33.2 m

_____________________________________________________________________________________________ 36. REASONING We will treat the horizontal and vertical

parts of the motion separately. The directions upward and to the right are chosen as the positive directions in the drawing. Ignoring air resistance, we can apply the equations of kinematics to the vertical part of the motion. The data for the vertical motion are summarized in the following tables. Note that the initial velocity component v0y is zero, because the bullets are

fired horizontally. The vertical component y of the bullets’ displacements are entered in the tables with minus signs, because the bullets move downward in the negative y direction. The times of flight tA and tB have been identified in the tables as a matter of convenience.

With these data, Equation 3.5b gives

( )1 1 12 2 20 2 2 2

0 m/sy y y yy v t a t t a t a t= + = + = (1) SOLUTION Applying Equation (1) to both shots, we find that

y-Direction Data, First Shot

y ay vy v0y t

−HA −9.80 m/s2 0 m/s tA

y-Direction Data, Second Shot

y ay vy v0y t

−HB −9.80 m/s2 0 m/s tB

v0x

x

y


1 2 2

BB B 2 B B1 22

A A A AA2

or y

y

a ty H H ty H H ta t

−= = =−

(2)

To use this result to calculate the ratio HB/HA, it is necessary to determine the times tA and

tB. To do this, we consider the horizontal part of the motion and note that there is no acceleration in the horizontal direction. Therefore, the horizontal component v0x of the bullet’s initial velocity remains unchanged throughout the motion, and the horizontal component x of the displacement is simply v0x times the time t during which the motion occurs. We have that

A B

A 0 A B 0 B A B0 0

and or and x xx x

x xx v t x v t t t

v v= = = = (3)

Substituting these results into Equation (2) gives

( )( )

2 22B 0B B B

2 2A AA A 0

/

/x

x

x vH t xH xt x v

⎛ ⎞= = = ⎜ ⎟⎜ ⎟⎝ ⎠

It is given that xB = 2xA, so that we find

2 2B B A

A A A

24

H x xH x x

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

37. SSM REASONING a. The drawing shows the initial velocity v0 of the

package when it is released. The initial speed of the package is 97.5 m/s. The component of its displacement along the ground is labeled as x. The data for the x direction are indicated in the data table below.


? 0 m/s2 +(97.5 m/s) cos 50.0° = +62.7 m/s

x

v0

y

50.0°

+x +y


Since only two variables are known, it is not possible to determine x from the data in this table. A value for a third variable is needed. We know that the time of flight t is the same for both the x and y motions, so let’s now look at the data in the y direction.

y-Direction Data

y ay vy v0y t

−732 m −9.80 m/s2 +(97.5 m/s) sin 50.0° = +74.7 m/s ? Note that the displacement y of the package points from its initial position toward the

ground, so its value is negative, i.e., y = −732 m. The data in this table, along with the appropriate equation of kinematics, can be used to find the time of flight t. This value for t can, in turn, be used in conjunction with the x-direction data to determine x.

b. The drawing at the right shows the velocity of the package

just before impact. The angle that the velocity makes with respect to the ground can be found from the inverse tangent function as ( )1tan /y xv vθ −= . Once the time has

been found in part (a), the values of vy and vx can be determined from the data in the tables and the appropriate equations of kinematics.

SOLUTION a. To determine the time that the package is in the air, we

will use Equation 3.5b ( )2120 y yy v t a t= + and the data in the y-direction data table. Solving

this quadratic equation for the time yields

( )( )( )

( ) ( ) ( )( )( )( )( )

2 120 0

12

2 212

212

4

2

74.7 m/s 74.7 m/s 4 9.80 m/s 732 m 6.78 s and 22.0 s

2 9.80 m/s

y y y

y

v v a yt

a

t

− ± − −=

− ± − −= = −

−

We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement x can be found using t = 22.0 s, the data in the x-direction data table, and Equation 3.5a:

x = v0x

t + 12 a

xt2 = +62.7 m/s( ) 22.0 s( ) + 1

2 0 m/s2( ) 22.0 s( )2

= 0

= +1380 m

θ

vx

+x +y

vy v


b. The angle θ that the velocity of the package makes with respect to the ground is given by

( )1tan /y xv vθ −= . Since there is no acceleration in the x direction (ax = 0 m/s2), vx is the

same as v0x, so that vx = v0x = +62.7 m/s. Equation 3.3b can be employed with the y-direction data to find vy :

( )( )20 74.7 m/s + 9.80 m/s 22.0 s 141 m/sy y yv v a t= + = + − = −

Therefore,

1 1 141 m/stan tan 66.062.7 m/s

y

x

vv

θ − −⎛ ⎞ −⎛ ⎞= = = − °⎜ ⎟ ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠

where the minus sign indicates that the angle is 66.0 below the horizontal° . ______________________________________________________________________________ 38. REASONING Since the vertical

height is asked for, we will begin with the vertical part of the motion, treating it separately from the horizontal part. The directions upward and to the right are chosen as the positive directions in the drawing. The data for the vertical motion are summarized in the following table. Note that the initial velocity component v0y is zero, because the marble is thrown horizontally. The vertical component y of the marble’s

displacement is entered in the table as −H, where H is the height we seek. The minus sign is included, because the marble moves downward in the negative y direction. The vertical component vy of the final velocity is checked as an important variable in the table, because we are given the angle that the final velocity makes

y-Direction Data

y ay vy v0y t

−H −9.80 m/s2 √ 0 m/s

with respect to the horizontal. Ignoring air resistance, we apply the equations of kinematics. With the data indicated in the table, Equation 3.6b becomes

( ) ( )2

22 20 2 0 m/s 2 or

2y

y y y yy

vv v a y a H H

a= + = + − = − (1)

65°

vx

vy

v0x

H

vy

vx


SOLUTION To use Equation (1), we need to determine the vertical component vy of the final velocity. We are given that the final velocity makes an angle of 65° with respect to the horizontal, as the inset in the drawing shows. Thus, from trigonometry, it follows that

0tan 65 or tan 65 or tan 65yy x y x

x

vv v v v

v

−° = = − ° = − °

where the minus sign is included, because vy points downward in the negative y direction.

In the absence of air resistance, there is no acceleration in the x direction, and the horizontal component vx of the final velocity is equal to the initial value v0x. Substituting this result into Equation (1) gives

( ) ( )( )

2220

2

15 m/s tan 65tan 6553 m

2 2 2 9.80 m/sy x

y y

v vH

a a⎡ ⎤− °− ° ⎣ ⎦= − = − = − =

−

39. REASONING As discussed in Conceptual Example 5, the horizontal velocity component

of the bullet does not change from its initial value and is equal to the horizontal velocity of the car. The same thing is true here for the tomato. In other words, regardless of its vertical position relative to the ground, the tomato always remains above you as you travel in the convertible. From the symmetry of free fall motion, we know that when you catch the tomato, its velocity will be 11 m/s straight downward. The time t required to catch the tomato can be found by solving Equation 3.3b ( 0y y yv v a t= + ) with 0–y yv v= . Once t is

known, the distance that the car moved can be found from xx v t= .

SOLUTION Taking upward as the positive direction, we find the flight time of the tomato to be

0 02

– –2 –2(11 m/s) 2.24 s–9.80 m/s

y y y

y y

v v vt

a a= = = =

Thus, the car moves through a distance of

(25 m/s )(2.24 s) 56 mxx v t= = = 40. REASONING The initial velocity of the diver has a

magnitude v0 and direction θ0 (the angle that the diver makes with respect to the horizontal). These two variables are shown in the drawing.

If we can determine the x and y components (v0x and

v0y) of her initial velocity, then the magnitude of her

θ0

v0

v0x

v0y

Water

+x

+y


initial velocity can be obtained from the Pythagorean

theorem ( )2 20 0 0x yv v v= + , and the angle θ0 can be

found by using the inverse tangent function 010

0tan y

x

vv

θ −⎡ ⎤⎛ ⎞=⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦.

SOLUTION The drawing at the right also shows the

velocity of the diver at the instant she contacts the water. Her speed is given as v = 8.90 m/s and her body is extended at an angle of θ = 75.0° with respect to the horizontal surface of the water.

We'll begin by examining the data for the x direction.

Since air resistance is being ignored, there is no acceleration in the x direction, so ax = 0 m/s2. Referring to the drawing, we see that the x component of the final velocity is given by

cos75.0 (8.90 m/s)cos75.0 2.30 m/sxv v= + ° = + ° = +

x-Direction Data

x ax vx v0x t

0 m/s2 +2.30 m/s ?

Since there is no acceleration in the x direction, the x component of the diver's velocity does not change. Thus, v0x = vx = +2.30 m/s.

To find a value for v0y we look at the data for the y direction. We are told that the vertical

displacement y of the diver is y = −3.00 m, where downward is chosen as the negative direction (see the drawing). The diver's acceleration in the y direction is that due to gravity, so ay = −9.80 m/s2. Finally, the y component of the diver's final velocity (see the drawing showing vy) is

sin75.0 (8.90 m/s)sin75.0 8.60 m/syv v= − ° = − ° = −

y-Direction Data

y ay vy v0y t

−3.00 m −9.80 m/s2 −8.60 m/s ?

v0

Water

v0x

v0y

+x

+y

vx

y

vy v

75.0°


Since three variables are known, we can use the relation 220 2y y yv v a y= + (Equation 3.6b)

to find v0y:

( ) ( )( )22 20 2 8.60 m/s 2 9.80 m/s 3.00 m 3.89 m/sy y yv v a y= + − = + − − − − = +

Since we now know values for v0x (+2.30 m/s) and v0y (+3.89 m/s), the Pythagorean

theorem can be used to determine the initial speed of the diver:

( ) ( )2 22 20 0 0 2.30 m/s 3.89 m/s 4.52 m/sx yv v v= + = + =

The angle θ0 can be found by using the inverse tangent function:

01 10

0

3.89 m/stan tan = 59.42.30 m/s

y

x

vv

θ − −⎛ ⎞ ⎛ ⎞= = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

______________________________________________________________________________ 41. REASONING The speed v of the soccer ball just before the goalie catches it is given by

2 2x yv v v= + , where vx and vx are the x and y components of the final velocity of the ball.

The data for this problem are (the +x direction is from the kicker to the goalie, and the +y direction is the “up” direction):

x-Direction Data

x ax vx v0x t

+16.8 m 0 m/s2 ? +(16.0 m/s) cos 28.0° = +14.1 m/s

y-Direction Data

y ay vy v0y t

−9.80 m/s2 ? +(16.0 m/s) sin 28.0° = +7.51 m/s

Since there is no acceleration in the x direction (ax = 0 m/s2), vx remains the same as v0x, so

vx = v0x = +14.1 m/s. The time t that the soccer ball is in the air can be found from the x-direction data, since three of the variables are known. With this value for the time and the y-direction data, the y component of the final velocity can be determined.


SOLUTION Since ax = 0 m/s2, the time can be calculated from Equation 3.5a as

0

16.8 m 1.19 s+14.1 m/sx

xtv

+= = = . The value for vy can now be found by using Equation 3.3b

with this value of the time and the y-direction data:

( )( )20 7.51 m/s 9.80 m/s 1.19 s 4.15 m/sy y yv v a t= + = + + − = −

The speed of the ball just as it reaches the goalie is

( ) ( )2 22 2 14.1 m/s 4.15 m/s 14.7 m/sx yv v v= + = + + − =

______________________________________________________________________________ 42. REASONING As shown in the drawing, the angle

that the velocity vector makes with the horizontal is given by

tan y

x

v

vθ =

where, from Equation 3.3b,

0 0 0siny y y yv v a t v a tθ= + = +

and, from Equation 3.3a (since ax = 0 m/s2),

0 0 0cosx xv v v θ= =

Therefore,

0 0

0 0

sintan

cosy y

x

v v a t

v v

θθ

θ+

= =

SOLUTION Solving for t, we find

( ) ( )( )0 0 0

2

cos tan – sin 29 m/s cos 36 tan 18 – sin 360.96 s

–9.80 m/sy

vt

aθ θ θ ° ° °

= = =

____________________________________________________________________________________________ 43. SSM REASONING The angle θ can be found from

1 2400 mtanx

θ − ⎛ ⎞= ⎜ ⎟⎝ ⎠ (1)


where x is the horizontal displacement of the flare. Since 0xa = m/s2, it follows that

0( cos 30.0 )x v t= ° . The flight time t is determined by the vertical motion. In particular, the time t can be found from Equation 3.5b. Once the time is known, x can be calculated.

SOLUTION From Equation 3.5b, assuming upward is the positive direction, we have

210 2–( sin 30.0 ) yy v t a t= ° +

which can be rearranged to give the following equation that is quadratic in t:

2102 ( sin 30.0 ) 0ya t v t y− ° − =

Using y = –2400 m and 29.80 m/sya = − and suppressing the units, we obtain the quadratic

equation 24.9 120 2400 0t t+ − =

Using the quadratic formula, we obtain t = 13 s. Therefore, we find that

x v t= ° = °( cos . ) ( ) (0 30 0 13 240 m/ s (cos 30.0 ) s) = 2700 m

Equation (1) then gives 1 2400 mtan 42

2700 mθ − ⎛ ⎞= = °⎜ ⎟⎝ ⎠

44. REASONING We can obtain an expression for the car’s initial velocity v0 by starting with

the relation 210 2x xx v t a t= + (Equation 3.5a). Here x is the horizontal component of the car’s

displacement, v0x is the horizontal component of the car’s initial velocity (v0x = v0 for a horizontal launch), and ax is its acceleration in the horizontal direction (ax = 0 m/s2 for projectile motion). Solving for the initial velocity, we find that

( )2 210 0 02 0 m/s or xx v t t v t v

t= + = =

(1)

The time t in Equation (1) is known, but x is not. To find x, we note that at t = 1.1 s, the car’s displacement has a magnitude of Δr = 7.0 m. The displacement Δr of the car has a horizontal component x and a vertical component y. The magnitude of the displacement is related to x and y by the Pythagorean theorem:

( )2 2 2r x yΔ = + (2)

Solving Equation (2) for x and substituting the result into Equation (1) gives


( )2 2

0r yxv

t tΔ −

= = (3) To determine y, we turn to the relation 21

0 2y yy v t a t= + (Equation 3.5b). Setting v0y = 0 m/s (again, for a horizontal launch), we find that Equation (3) becomes

( ) 2 21 1

2 20 m/s y yy t a t a t= + = (4)

SOLUTION Substituting Equation (4) into Equation (3) gives

( ) ( )2 2 42 124

0yr a tr y

vt t

Δ −Δ −= =

Thus, the car’s speed just as it drives off the end of the dock is

( ) ( ) ( )22 4214

0

7.0 m 9.80 m/s 1.1 s3.4 m/s

1.1 sv

− −= =

45. REASONING The following drawings show the initial and final velocities of the ski

jumper and their scalar components. The initial speed of the ski jumper is given

by 2 20 0 0x yv v v= + , and the angle that the initial velocity makes with the horizontal is

01

0

tan y

x

vv

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

. The scalar components v0x and v0y can be determined by using the

equations of kinematics and the data in the following tables. (The +x direction is in the direction of the horizontal displacement of the skier, and the +y direction is “up.”)

x-Direction Data

x ax vx v0x t

+51.0 m 0 m/s2 +(23.0 m/s) cos 43.0° = +16.8 m/s ?

y-Direction Data

θ v0x

v0y v0

Initial velocity Final velocity

43.0° vy

vx

v


y ay vy v0y t

−9.80 m/s2 −(23.0 m/s) sin 43.0° = −15.7 m/s ?

Since there is no acceleration in the x direction (ax = 0 m/s2), v0x is the same as vx, so we

have that v0x = vx = +16.8 m/s. The time that the skier is in the air can be found from the x-direction data, since three of the variables are known. With the value for the time and the y-direction data, the y component of the initial velocity can be determined.

SOLUTION Since ax = 0 m/s2, the time can be determined from Equation 3.5a as

0

51.0 m 3.04 s+16.8 m/sx

xtv

+= = = . The value for v0y can now be found by using Equation 3.3b

with this value of the time and the y-direction data:

( )( )20 15.7 m/s 9.80 m/s 3.04 s 14.1 m/sy y yv v a t= − = − − − = +

The speed of the skier when he leaves the end of the ramp is

( ) ( )2 22 20 0 0 16.8 m/s 14.1 m/s 21.9 m/sx yv v v= + = + + + =

The angle that the initial velocity makes with respect to the horizontal is

01 1

0

14.1 m/stan tan 40.016.8 m/s

y

x

vv

θ − −⎛ ⎞ +⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ +⎝ ⎠⎝ ⎠

______________________________________________________________________________

46. REASONING The height of each building is the magnitude of the vertical displacement y of the stone that is thrown from it. We will determine y from 1 2

0 2y yy v t a t= +

(Equation 3.5b). We choose this equation because we know that 0 0.00 m / syv = (the

stones initially travel horizontally) and that 29.80 m/sya = − (assuming that upward is the +y direction). We do not know the value of the fall time t for either stone. However, we can relate the fall time to the horizontal displacement x that each stone has when it lands. This is possible because the motion of the stones in the horizontal direction occurs at a constant velocity 0xv , so that we can use 0xx v t= to determine the displacement x.

SOLUTION From Equation 3.5b we have that

( )1 1 12 2 20 2 2 2

0.00 m / sy y y yy v t a t t a t a t= + = + =


The motion in the horizontal direction occurs at a constant velocity of 0xv , so that the horizontal displacement x and the fall time t can be related as follows:

00

or xx

xx v t tv

= =

Substituting this result for t into the expression for y gives 2 2

1 1222 2

0 02y

y yx x

a xxy a t av v

⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠

This expression for y applies to each stone with different values for y and x but the same values for ay and v0x. Thus, the ratio ytaller/yshorter is

2taller2 2

taller 0 taller2 2

shorter shorter shorter20

2

2

y

x

y

x

a x

y v xy a x x

v

= =

We also know that taller shorter2x x= , since the stone thrown from the top of the taller building lands twice as far from the base of the building as does the other stone. Therefore, we find that

( )22shortertaller taller

2 2shorter shorter shorter

24

xy xy x x

= = =

We see, then, that the ratio of the height of the taller building to the height of the shorter

building is 4 . ____________________________________________________________________________________________ 47. SSM REASONING AND SOLUTION In the absence of air resistance, the bullet

exhibits projectile motion. The x component of the motion has zero acceleration while the y component of the motion is subject to the acceleration due to gravity. The horizontal distance traveled by the bullet is given by Equation 3.5a (with 0xa = m/s2):

0 0( cos )xx v t v tθ= =

with t equal to the time required for the bullet to reach the target. The time t can be found

by considering the vertical motion. From Equation 3.3b,

0y y yv v a t= +


When the bullet reaches the target, 0y yv v= − . Assuming that up and to the right are the positive directions, we have

0 0 00

2 2 sin 2 sinand ( cos )y

y y y

v v vt x v

a a aθ θ

θ⎛ ⎞− − −⎜ ⎟= = =⎜ ⎟⎝ ⎠

Using the fact that 2sin cos sin 2θ θ θ= , we have

2 20 02 cos sin sin 2

y y

v vx

a aθ θ θ

= − = −

Thus, we find that

23

2 20

(91.4 m) (–9.80 m/s )sin 2 4.91 10(427 m/s)

yx a

vθ −= − = − = ×

and

2 0.281 or 2 180.000 0.281 179.719θ θ= ° = °− ° = ° Therefore,

0.141 and 89.860θ = ° ° _____________________________________________________________________________________________ 48. REASONING We will treat the horizontal and vertical parts of the motion separately. The

range R is the product of the horizontal component of the initial velocity v0x and the time of flight t. The time of flight can be obtained from the vertical part of the motion by using Equation 3.5b

y = v

0 yt + 1

2 ayt2( ) and the fact that the displacement y in the vertical direction

is zero, since the projectile is launched from and returns to ground level. The expression for the range obtained in this way can then be applied to obtain the desired launch angle for doubling the range.

SOLUTION The range of the projectile is

Using Equation 3.5b, we obtain the time of flight as

y v t a t tv

avay y

y

y y

= = + = − = −02 2

012

2 0 0 or sinθ

Substituting this expression for t into the range expression gives

R = v0 cosθ( ) −2v0 sinθ

ay

⎛

⎝⎜⎜

⎞

⎠⎟⎟= −

2v02 cosθ sinθ

ay= −

v02 sin 2θ

ay

R = v0x

t = v0cosθ t


where we have used the fact that 2 cos θ sin θ = sin 2θ. We can now apply this expression for the range to the initial range R1 for θ1 and the range R2 for θ2:

Rva

Rv

ay y1

02

12

02

22 2= − = −

sin sinθ θ or

Dividing these two expressions gives

R2R1

=− v0

2 sin 2θ2( ) / ay− v0

2 sin 2θ1( ) / ay= sin 2θ2sin 2θ1

= 2

where we have used the fact that R2/R1 = 2. Since θ1 = 12.0°, we find that

sin 2θ2 = 2.00sin 2θ1 = 2.00sin 2 12.0°( ) = 0.813

θ2 =sin−1 0.8132.00

= 27.2°

49. SSM REASONING Since the horizontal motion is not accelerated, we know that the x

component of the velocity remains constant at 340 m/s. Thus, we can use Equation 3.5a (with 0xa = m/s2) to determine the time that the bullet spends in the building before it is embedded in the wall. Since we know the vertical displacement of the bullet after it enters the building, we can use the flight time in the building and Equation 3.5b to find the y component of the velocity of the bullet as it enters the window. Then, Equation 3.6b can be used (with 0 0yv = m/s) to determine the vertical displacement y of the bullet as it passes between the buildings. We can determine the distance H by adding the magnitude of y to the vertical distance of 0.50 m within the building.

Once we know the vertical displacement of the bullet as it passes between the buildings, we

can determine the time 1t required for the bullet to reach the window using Equation 3.4b. Since the motion in the x direction is not accelerated, the distance D can then be found from

0 1xD v t= . SOLUTION Assuming that the direction to the right is positive, we find that the time that

the bullet spends in the building is (according to Equation 3.5a)

0

6.9 m 0.0203 s340 m/sx

xtv

= = =


The vertical displacement of the bullet after it enters the building is, taking down as the negative direction, equal to –0.50 m. Therefore, the vertical component of the velocity of the bullet as it passes through the window is, from Equation 3.5b,

1 21 1 22

0 (window) 2 2–0.50 m ( 9.80 m/s )(0.0203 s) = –24.5 m/s0.0203 s

yy y

y a t yv a tt t

−= = − = − −

The vertical displacement of the bullet as it travels between the buildings is (according to

Equation 3.6b with 0 0yv = m/s)

2 2

2( 24.5 m/s) 30.6 m

2 2(–9.80 m/s )y

y

vy

a−= = = −

Therefore, the distance H is

30.6 m 0.50 m 31 mH = + =

The time for the bullet to reach the window, according to Equation 3.4b, is

10

2 2 2(–30.6 m) 2.50 s(–24.5 m/s)y y y

y ytv v v

= = = =+

Hence, the distance D is given by

0 1 (340 m/s)(2.50 s) 850 mxD v t= = = 50. REASONING The angle θ is the angle that the balloon’s initial velocity v0 makes with the

horizontal, and can be found from the horizontal and vertical components of the initial velocity:

0 01

0 0tan or tany y

x x

v v

v vθ θ − ⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠

(1)

(Note: because the balloon follows a curved trajectory, θ is not related in this fashion to the horizontal and vertical components x and y of the balloon’s displacement.) We will use

210 2x xx v t a t= + (Equation 3.5a) and 21

0 2y yy v t a t= + (Equation 3.5b) to find expressions for the horizontal and vertical components (v0x and v0y) of the balloon’s initial velocity, and then combine those results with Equation (1) to find θ. For the initial horizontal velocity component, with ax = 0 m/s2 since air resistance is being ignored, Equation 3.5a gives


( )2 210 0 02 0 m/s or x x x

xx v t t v t vt

= + = = (2)

For the initial vertical velocity component, we obtain from Equation 3.5b that

21

2210 02 or yy y y

y a ty v t a t v

t−

= + = (3)

In part b, the initial speed v0 of the second balloon can be found by first noting that it is related to the x component v0x of the balloon’s initial velocity by 0 0 cosxv v θ= . Since there is no acceleration in the x direction (ax = 0 m/s2), v0x is equal to the x component of the balloon’s displacement (x = +35.0 m) divided by the time t that the second balloon is in the air, or v0x = x/t. Thus, the relation 0 0 cosxv v θ= can be written as

0 cosx vt

θ= (4)

In Equation (4), x = +35.0 m and θ is known from the result of part (a). The time of flight t is related to the y component of the balloon’s displacement by 21

0 2y yy v t a t= + (Equation

3.5b). Since 0 0 sinyv v θ= , Equation 3.5b can be expressed as

( )2 21 10 02 2siny y yy v t a t v t a tθ= + = + (5)

Equations (4) and (5) provide everything necessary to find the initial speed v0 of the water balloon for the second launch. SOLUTION a. Substituting Equations (2) for v0x and (3) for v0y into Equation (1) gives an expression for the balloon’s initial direction θ :

212

01 1

0tan tan

y

y

x

y a tv tv

θ − −

−⎛ ⎞

= =⎜ ⎟⎝ ⎠ x

t

2121tan yy a tx

−

⎡ ⎤⎢ ⎥ ⎡ ⎤−⎢ ⎥ ⎢ ⎥=⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

(5)

The balloon leaves the roof of Jackson, 15.0 m above the ground, and hits halfway up Walton, a point 12 (22.0 m) = 11.0 m above the ground. The balloon’s height at impact is therefore 15.0 m − 11.0 m = 4.0 m below its launch height. Taking up as the positive direction, the vertical displacement of the balloon is therefore y = −4.0 m, and its horizontal


displacement is x = +35.0 m, the distance between the buildings. Therefore, the angle at which the first balloon is launched is

( )( )22121

4.0 m 9.80 m/s 2.0 stan 24

35.0 mθ −

⎡ ⎤− − −⎢ ⎥= =⎢ ⎥⎣ ⎦

o

b. The angle θ is that found in part (a), and the vertical displacement is the difference in heights between the Walton & Jackson dorms: y = 22.0 m − 15.0 m = +7.0 m. Solving Equation (4) for the elapsed time t and substituting the result into Equation (5) gives

( ) ( )2

21 10 02 2

0 0sin sin

cos cosy yx xy v t a t v a

v vθ θ

θ θ⎛ ⎞ ⎛ ⎞

= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Solving this equation for the initial speed v0 of the second balloon gives

( )( )

21122

0

m/s35.0 m 29 m/ssin sin 24cos cos24 7.0 m 35.0 mcos cos24

yaxvy x θθ

θ

−9.80= = =°°− −

°

51. REASONING We can use information about the motion of clown A and the collision to

determine the initial velocity components for clown B. Once the initial velocity components are known, the launch speed 0Bv and the launch angle Bθ for clown B can be determined.

SOLUTION From Equation 3.5b y = v0yt +

12 ayt

2( ) we can find the time of flight until the collision. Taking upward as positive and noting for clown A that v0y = (9.00 m/s) sin 75.0° = 8.693 m/s, we have

1.00 m= 8.693 m/s( )t + 1

2 –9.80 m/s2( )t 2

Rearranging this result and suppressing the units gives

24.90 –8.693 1.00 0t t + = The quadratic equation reveals that

t =8.693± –8.693( )2 – 4 4.90( ) 1.00( )

2 4.90( ) = 1.650 s or 0.1237 s


Using these values for t with the magnitudes v0xA and v0xB of the horizontal velocity components for each clown, we note that the horizontal distances traveled by each clown must add up to 6.00 m. Thus,

0 0 0 06.00 m6.00 m or –xA xB xB xAv t v t v vt

+ = =

Using v0xA = (9.00 m/s) cos 75.0° = 2.329 m/s, we find

0 06.00 m 6.00 m– 2.329 m/s 1.307 m/s or – 2.329 m/s 46.175 m/s1.650 s 0.1237 sxB xBv v= = = =

The vertical component of clown B’s velocity is v0yB and must be the same as that for

clown A, since each clown has the same vertical displacement of 1.00 m at the same time. Hence, v0yB = 8.693 m/s (see above). The launch speed of clown B, finally, is

v v vB xB yB0 02

02= + . Thus, we find

v0B = 1.307 m/s( )2 + 8.693 m/s( )2 = 8.79 m/s or

v0B = 46.175 m/s( )2 + 8.693 m/s( )2 = 47.0 m/s For these two possible launch speeds, we find the corresponding launch angles using the

following drawings, neither of which is to scale:

8.693 m/sv0B

1.307 m/s

θB

+x

+y

8.693 m/sv0B

46.175 m/s

θB+x

+y

–1 –18.693 m/s 8.693 m/stan 81.5 tan 10.71.307 m/s 46.175 m/s

θ θ⎛ ⎞ ⎛ ⎞= = ° = = °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Since the problem states that θB > 45°, the solution is 0 B8.79 m/s and 81.5Bv θ= = ° .

52. REASONING The time it takes for John to pass Chad is the distance between them divided by John’s speed vJC relative to Chad. We need, then to find John’s velocity relative to Chad.


SOLUTION In computing the necessary relative velocity, we define the following symbols:

vJC = John’s velocity relative to Chad vJG = John’s velocity relative to the ground vGC = The ground’s velocity relative to Chad vCG = Chad’s velocity relative to the ground

John’s velocity relative to Chad is

vJC = vJG + vGC = vJG − vCG = 4.50 m/s, north( )− 4.00 m/s, north( ) = 0.50 m/s, north Here we have made use of the fact that vGC = –vCG. The time it takes for John to pass Chad

is

JC

95 m 95 m 190 s0.50 m/s

tv

= = =

______________________________________________________________________________ 53. SSM REASONING The velocity SGv of the

swimmer relative to the ground is the vector sum of the velocity SWv of the swimmer relative to the water and the velocity WGv of the water relative to the ground as shown at the right: SG SW WG= +v v v .

The component of SGv that is parallel to the

width of the river determines how fast the swimmer is moving across the river; this parallel

component is SWv . The time for the swimmer to cross the river is equal to the width of the river divided by the magnitude of this velocity component.

The component of SGv that is parallel to the direction of the current determines how far the

swimmer is carried down stream; this component is WGv . Since the motion occurs with constant velocity, the distance that the swimmer is carried downstream while crossing the river is equal to the magnitude of WGv multiplied by the time it takes for the swimmer to cross the river.

SOLUTION a. The time t for the swimmer to cross the river is


33

SW

width 2.8 10 m 2.0 10 s1.4 m/s

tv

×= = = ×

b. The distance x that the swimmer is carried downstream while crossing the river is

3 3WG (0.91 m/s)(2.0 10 s) 1.8 10 mx v t= = × = ×

54. REASONING There are three velocities involved:

NBv = velocity of Neil relative to Barbara

ΝGv = velocity of Neil relative to the Ground (3.2 m/s, due west)

BGv = velocity of Barbara relative to the Ground (4.0 m/s, due south)

Ordering the vectors vNB , vNG and vGB by their subscripts in the manner discussed in Section 3.4 of the text, we see that = +NB NG GBv v v . Note that this ordering involves GBv , the velocity of the ground relative to Barbara. According to the discussion in Section 3.4, GBv is related to BGv by GBv = − BGv . Thus, the three vectors listed above are related by

( )= + −NB NG BGv v v . The drawing shows this relationship. Since the three vectors form a right triangle, the magnitudes of the vectors are related by the Pythagorean theorem, which we will use to obtain Neil’s speed relative to Barbara. Trigonometry can be used to determine the angle θ.

SOLUTION Using the Pythagorean theorem, we find that the speed vNB of Neil relative to Barbara is

( ) ( ) ( )2 2 22NB NG BG 3.2 m/s 4.0 m/s 5.1 m/sv v v= + − = + − =

The angle θ can be found from trigonometry:

( )NG

NB

3.2 m/scos cos 51 north of west5.1 m/s

vv

θ −1 −1⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

55. REASONING There are three velocities involved:

North

West θ

− BGv

NGv

NBv


TGv = the velocity of the Truck relative to the Ground

TPv = the velocity of the Truck relative to the Police car

PGv = the velocity of the Police car relative to the Ground

Of these three vectors, we know the directions of PGv (north) and TGv (west), and the magnitudes of PGv (29 m/s) and TPv (48 m/s). Ordering these vectors by their subscripts in the manner discussed in Section 3.4 of the text, we have = +TG TP PGv v v . The triangle

formed by this vector sum is a right triangle, because two of the vectors ( ) and TG PGv v are perpendicular to one another. The third vector, TPv , must be the triangle’s hypotenuse. The magnitudes of the three vectors are thus related to one another by the Pythagorean theorem, which we will use to obtain the truck’s speed relative to the ground.

SOLUTION a. To represent the vector sum = +TG TP PGv v v graphically, the vectors being added together, TPv and PGv , must be drawn tail-to-head (see the drawing). The resultant vector

TGv runs from the tail of TPv to the head of PGv . The direction of TPv shows that, relative to the police car, the truck is moving both west and south.

b. As the drawing shows, TPv is the hypotenuse of the vector right triangle. Therefore, the Pythagorean theorem gives

2 2 2 2 2TP TG PG TG TP PG or v v v v v v= + = −

The speed of the truck relative to the ground is

( ) ( )2 22 2TG TP PG 48 m/s 29 m/s 38 m/sv v v= − = − =

This is equivalent to 85 mph.

56. REASONING When you use the speed ramp, the time it takes you to cover the ground

distance 105 md = is the distance divided by your speed relative to the ground. To determine your speed relative to the ground we will refer to the following relative velocities:

vYG = Your velocity relative to the Ground vYR = Your velocity relative to the speed Ramp vRG = The speed Ramp’s velocity relative to the Ground

SOLUTION While you are on the ramp, your speed vYG relative to the ground is the

magnitude of the velocity vYG. The desired time t is

TPv PGv

North

West

vTG


YG

dtv

=

Using the relative velocities defined in the REASONING, your velocity relative to the

ground is given by

= +YG YR RGv v v

Since you walk with respect to the ramp at the same rate that you walk on the ground, your

velocity vYR with respect to the ramp has a magnitude of YR105 m 1.4 m / s75 s

v = = .

Moreover, we know that the magnitude of the ramp’s velocity relative to the ground is RG 2.0 m / sv = . Since you and the ramp are moving in the same direction, the velocities vYR

and vRG have the same direction (assumed to be the positive direction), and we can write that

( ) ( )1.4 m / s 2.0 m / s 3.4 m / s= + = + + + = +YG YR RGv v v

Thus, using our expression for the time, we obtain

YG

105 m 31 s3.4 m / s

dtv

= = =

______________________________________________________________________________ 57. REASONING Let HBv represent the velocity of the hawk relative to the balloon and BGv

represent the velocity of the balloon relative to the ground. Then, as indicated by Equation 3.7, the velocity of the hawk relative to the ground is HG HB BG= +v v v . Since the vectors HBv and BGv are at right angles to each other, the vector addition can be carried out using the Pythagorean theorem.

SOLUTION Using the drawing at the right, we have

from the Pythagorean theorem,

v v vHG HB2

BG2

2 2 (2.0 m / s) (6.0 m / s) 6.3 m / s

= +

= + =

The angle θ is

vHBθ

EAST

vHG

vBG

NORTH

θ = tan−1 vHB

vBG

⎛⎝⎜

⎞⎠⎟= tan−1 2.0 m/s

6.0 m/s⎛⎝⎜

⎞⎠⎟ = 18°, north of east


58. REASONING The time it takes for the passenger to walk the distance on the boat is the distance divided by the passenger’s speed vPB relative to the boat. The time it takes for the passenger to cover the distance on the water is the distance divided by the passenger’s speed vPW relative to the water. The passenger’s velocity relative to the boat is given. However, we need to determine the passenger’s velocity relative to the water.

SOLUTION a. In determining the velocity of the passenger relative to the water, we define

the following symbols:

vPW = Passenger’s velocity relative to the water vPB = Passenger’s velocity relative to the boat vBW = Boat’s velocity relative to the water

The passenger’s velocity relative to the water is

vPW = vPB + vBW = 1.5 m/s, north( ) + 5.0 m/s, south( ) = 3.5 m/s, south

b. The time it takes for the passenger to walk a distance of 27 m on the boat is

PB

27 m 27 m 18 s1.5 m/s

tv

= = =

c. The time it takes for the passenger to cover a distance of 27 m on the water is

PW

27 m 27 m 7.7 s3.5 m/s

tv

= = =

____________________________________________________________________________________________ 59. SSM REASONING The velocity ABv of train A relative to train B is the vector sum of

the velocity AGv of train A relative to the ground and the velocity GBv of the ground relative to train B, as indicated by Equation 3.7: AB AG GB= +v v v . The values of AGv and

BGv are given in the statement of the problem. We must also make use of the fact that GB BG= −v v .

SOLUTION a. Taking east as the positive direction, the velocity of A relative to B is, according to

Equation 3.7,

AB AG GB AG BG (+13 m/s) – (–28 m/s) +41 m/s= + = − = =v v v v v

The positive sign indicates that the direction of ABv is due east . b. Similarly, the velocity of B relative to A is


BA BG GA BG AG (–28 m/s) – (+13 m/s) –41 m/s= + = − = =v v v v v

The negative sign indicates that the direction of BAv is due west . 60. REASONING Using subscripts to make the relationships among the three relative

velocities clear, we have the following: vPG = the velocity of the Plane relative to the Ground (unknown speed, due west) vPA = the velocity of the Plane relative to the Air (245 m/s, unknown direction) vAG = the velocity of the Air relative to the Ground (38.0 m/s, due north)

Arranging the subscripts as shown in Section 3.4 of the text, we find that the velocity of the plane relative to the ground is the vector sum of the other two velocities: vPG = vPA + vAG. This vector sum may be illustrated as follows:

As the three relative velocity vectors form a right triangle, we will apply trigonometry to find the direction θ that the pilot should aim the plane relative to due west.

SOLUTION The magnitudes of the two known sides of the vector right triangle are vPA (245 m/s) and vAG (38.0 m/s) which are, respectively, the hypotenuse and the side opposite the angle θ (see the drawing). The sine of the angle θ is therefore the ratio of vAG to vPA, so that the direction the pilot should head the plane is

1 1AG

PA

38.0 m/ssin sin 8.92 , south of west245 m/s

vv

θ − −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

o

61. REASONING AND SOLUTION The velocity of the raindrops relative to the train is given by

vRT = vRG + vGT

where vRG is the velocity of the raindrops relative to the ground and vGT is the velocity of the ground relative to the train.

North vPG

vAG vPA θ East


Since the train moves horizontally, and the rain falls

vertically, the velocity vectors are related as shown in the figure at the right. Then

vGT = vRG tan θ = (5.0 m/s) (tan 25°) = 2.3 m/s The train is moving at a speed of 2.3 m/s

vRG

GTv

RTv25°

____________________________________________________________________________________________ 62. REASONING The relative velocities in this problem are: vPW = velocity of the Passenger relative to the Water vPB = velocity of the Passenger relative to the Boat (2.50 m/s, due east) vBW = velocity of the Boat relative to the Water (5.50 m/s, at 38.0° north of east)

The velocities are shown in the drawing are related by the subscripting method discussed in

Section 3.4: vPW = vPB + vBW

We will determine the magnitude and direction of vPW from the equation above by using the

method of scalar components. SOLUTION The table below lists the scalar components of the three vectors.

Vector x Component y Component

vPB +2.50 m/s 0 m/s

vBW +(5.50 m/s) cos 38.0° = +4.33 m/s +(5.50 m/s) sin 38.0° = +3.39 m/s

vPW = vPB + vBW +2.50 m/s + 4.33 m/s = +6.83 m/s 0 m/s + 3.39 m/s = +3.39 m/s

The magnitude of vPW can be found by applying the Pythagorean theorem to the x and y components:

( ) ( )2 2PW 6.83 m/s 3.39 m/s 7.63 m/sv = + =

The angle θ (see the drawings) that vPW makes with due east is

vPB

vBW

vPW

θ

5.50 m/s

vPW

θ

2.50 m/s 38.0°

+x (East)

+y (North)


1 3.39 m/stan 26.4 north of east6.83 m/s

θ − +⎛ ⎞= = °⎜ ⎟+⎝ ⎠

______________________________________________________________________________ 63. SSM REASONING The velocity PMv of the puck relative to Mario is the vector sum of

the velocity PIv of the puck relative to the ice and the velocity IMv of the ice relative to Mario as indicated by Equation 3.7: PM PI IM= +v v v . The values of MIv and PIv are given in the statement of the problem. In order to use the data, we must make use of the fact that IM MI= −v v , with the result that PM PI MI= −v v v .

SOLUTION The first two rows of the following table give the east/west and north/south

components of the vectors PIv and MI–v . The third row gives the components of their resultant PM PI MI= −v v v . Due east and due north have been taken as positive.

Vector

East/West Component

North/South Component PIv

–(11.0 m/s) sin 22° = –4.1 m/s

–(11.0 m/s) cos 22° = –10.2 m/s

MI–v

0

+7.0 m/s

PM PI MI–=v v v

–4.1 m/s

–3.2 m/s

Now that the components of PMv are known, the

Pythagorean theorem can be used to find the magnitude.

vPM2 2(–4.1 m / s) / s) 5.2 m / s= + =(–3.2 m

The direction of PMv is found from

φ = tan−1 4.1 m/s3.2 m/s

⎛⎝⎜

⎞⎠⎟ = 52° west of south

64. REASONING AND SOLUTION While flying west, the airplane has a ground speed of

vPG = 2.40 × 102 m/s – 57.8 m/s = 182 m/s

and requires time tW = x/(182 m/s) to reach the turn-around point. While flying east the

airplane has a ground speed of


vPG = 2.40 × 102 m/s + 57.8 m/s = 298 m/s

and requires tE = x/(298 m/s) to return home. Now the total time for the trip is

t = tW + tE = 6.00 h = 2.16 × 104 s, so

x/182 + x/298 = 2.16 × 104 or x = 62.44 10 m 2440 km× = 65. SSM REASONING The relative velocities in this problem are:

vPS = velocity of the Passenger relative to the Shore vP2 = velocity of the Passenger relative to Boat 2 (1.20 m/s, due east) v2S = velocity of Boat 2 relative to the Shore v21 = velocity of Boat 2 relative to Boat 1 (1.60 m/s, at 30.0° north of east) v1S = velocity of Boat 1 relative to the Shore (3.00 m/s, due north)

The velocity vPS of the passenger relative to the shore is related to vP2 and v2S by (see the

method of subscripting discussed in Section 3.4):

vPS = vP2 + v2S But v2S, the velocity of Boat 2 relative to the

shore, is related to v21 and v1S by

v2S = v21 + v1S Substituting this expression for v2S into the

first equation yields vPS = vP2 + v21 + v1S

This vector sum is shown in the diagram. We

will determine the magnitude of vPS from the equation above by using the method of scalar components.

SOLUTION The table below lists the scalar components of the four vectors in the drawing.

Vector x Component y Component

vP2 +1.20 m/s 0 m/s

v21 +(1.60 m/s) cos 30.0° = +1.39 m/s +(1.60 m/s) sin 30.0° = +0.80 m/s

vP2

v21

vPS

1.20 m/s +x (East)

+y (North) v1S

3.00 m/s

30.0°

1.60 m/s


v1S 0 m/s +3.00 m/s

vPS = vP2 + v21 + v1S +1.20 m/s + 1.39 m/s = +2.59 m/s +0.80 m/s +3.00 m/s = +3.80 m/s

The magnitude of vPS can be found by applying the Pythagorean theorem to its x and y

components:

( ) ( )2 2PS 2.59 m/s 3.80 m/s 4.60 m/sv = + =

______________________________________________________________________________ 66. REASONING The magnitude and direction of the initial velocity v0 can be obtained using

the Pythagorean theorem and trigonometry, once the x and y components of the initial velocity v0x and v0y are known. These components can be calculated using Equations 3.3a and 3.3b.

SOLUTION Using Equations 3.3a and 3.3b, we obtain the following results for the velocity

components:

v0x = vx − axt = 3775 m/s− 5.10 m/s2( ) 565 s( ) = 893.5 m/s

v0y = vy − ayt = 4816 m/s− 7.30 m/s2( ) 565 s( ) = 691.5 m/s

Using the Pythagorean theorem and trigonometry, we find

v0 = v0x2 + v0y

2 = 893.5 m/s( )2 + 691.5 m/s( )2 = 1130 m/s

θ = tan−1 v0y

v0x

⎛⎝⎜

⎞⎠⎟= tan−1 691.5 m/s

893.5 m/s⎛⎝⎜

⎞⎠⎟ = 37.7°

____________________________________________________________________________________________ 67. REASONING Trigonometry indicates that the x and y components of the dolphin’s velocity

are related to the launch angle θ according to tan θ = vy /vx. SOLUTION Using trigonometry, we find that the y component of the dolphin’s velocity is

vy = vx tanθ = vx tan 35° = 7.7 m/s( ) tan 35° = 5.4 m/s 68. REASONING The upward direction is chosen as positive. Since the ballast bag is released

from rest relative to the balloon, its initial velocity relative to the ground is equal to the


velocity of the balloon relative to the ground, so that 0 3.0 m/syv = . Time required for the ballast to reach the ground can be found by solving Equation 3.5b for t.

SOLUTION Using Equation 3.5b, we have

12

20 0a t v t yy y+ − = or 1

22 3 0 9 5 0(–9.80 m / s m / s) m)2 ) ( . ( .t t+ − − =

This equation is quadratic in t, and t may be found from the quadratic formula. Using the

quadratic formula, suppressing the units, and discarding the negative root, we find

t =− ± − −

−=

3 0 4 4 90 9 52 4 90

. ( . )( . )( . )

(3.0)1.7 s

2

____________________________________________________________________________________________ 69. REASONING The time that the ball spends in the air is determined by its vertical motion.

The time required for the ball to reach the lake can be found by solving Equation 3.5b for t. The motion of the golf ball is characterized by constant velocity in the x direction and accelerated motion (due to gravity) in the y direction. Thus, the x component of the velocity of the golf ball is constant, and the y component of the velocity at any time t can be found from Equation 3.3b. Once the x and y components of the velocity are known for a particular time t, the speed can be obtained from v v vx y= +2 2 .

SOLUTION a. Since the ball rolls off the cliff horizontally, v0y = 0. If the origin is chosen at top of the

cliff and upward is assumed to be the positive direction, then the vertical component of the ball's displacement is y = – 15.5 m. Thus, Equation 3.5b gives

tya y

= = − =2 2(

)15.5 m)

(–9.80 m / s1.78 s2

b. Since there is no acceleration in the x direction, 0 11.4 m/sx xv v= = . The y component of

the velocity of the ball just before it strikes the water is, according to Equation 3.3b,

20 0 + (–9.80 m/s )(1.78 s) –17.4 m/sy y yv v a t ⎡ ⎤= + = =⎣ ⎦

The speed of the ball just before it strikes the water is, therefore,

v v vx y= + = + − =2 2 (11.4 m / s) 17.4 m / s) 20.8 m / s2 2( ____________________________________________________________________________________________


70. REASONING As shown in Example 8 in the text, the range R of a projectile is given by R = v0xt, where 0 0 cosxv v θ= is the horizontal component of the launch velocity of magnitude v0 and launch angle θ, and t is the total flight time of the projectile. In the absence of air resistance, the time required for the projectile to rise is equal to the time required for the projectile to fall; therefore, the total flight time t is equal to twice the time tH required for a projectile to reach its maximum vertical displacement H. The time tH can be found from 0y y yv v a t= + (Equation 3.3b) by setting 0yv = and solving for t = tH.

SOLUTION a. With upward taken as the positive direction, the time tH for the greyhound to reach its

maximum vertical displacement H is given by Equation 3.3b as

( )0 0H 2

sin 31.0 10.0 m/s sin 31.0– – – 0.526 s

–9.80 m/sy

y y

v vt

a a° °

= = = =

The range of the leap is, therefore,

( ) ( )0 0( cos 31.0 ) 10.0 m/s cos 31.0 2 0.526 s = 9.02 mxR v t v t= = ° = °⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

b. Ignoring air resistance, we know that the time for the projectile to rise from ground level

to its maximum height is equal to the time for the projectile to fall back to ground level. Therefore, the greyhound is in the air for ( )2 0.526 s 1.05 s= .

_____________________________________________________________________________________________ 71. SSM REASONING Once the diver is airborne, he moves in the x direction with constant

velocity while his motion in the y direction is accelerated (at the acceleration due to gravity). Therefore, the magnitude of the x component of his velocity remains constant at 1.20 m/s for all times t. The magnitude of the y component of the diver's velocity after he has fallen through a vertical displacement y can be determined from Equation 3.6b: 2 2

0 2y y yv v a y= + .

Since the diver runs off the platform horizontally, 0 0yv = m/s. Once the x and y components of the velocity are known for a particular vertical displacement y, the speed of the diver can be obtained from v v vx y= +2 2 .

SOLUTION For convenience, we will take downward as the positive y direction. After the

diver has fallen 10.0 m, the y component of his velocity is, from Equation 3.6b,

v v a yy y y= + = =02 2 10 0 14 00 + 2(9.80 m / s m) m / s2 2 )( . .

Therefore,

v v vx y= + = + =2 2 (1.20 m / s) 14.0 m / s) 14.1 m / s2 2(


____________________________________________________________________________________________ 72. REASONING When the ball is thrown straight up with an initial speed v0, the maximum

height y that it reaches can be found by using with the relation 2 20 2y y yv v a y= + (Equation

3.6b). Since the ball is thrown straight up, v0y = v0, where v0 is the initial speed of the ball. Also, the speed of the ball is momentarily zero at its maximum height, so vy = 0 m/s at that point. The acceleration ay is that due to gravity, so the only unknown besides y is the initial speed v0 of the ball. To determine v0 we will employ Equation 3.6b a second time, but now it will be applied to the case where the ball is thrown upward at an angle of 52º above the horizontal. In this case the maximum height reached by the ball is y1 = 7.5 m, the initial speed in the y direction is v0y = v0 sin 52º, and the y-component of the speed at the maximum height is vy = 0 m/s.

SOLUTION We will start with the relation 2 2

0 2y y yv v a y= + (Equation 3.6b) to find the maximum height y that the ball attains when it is thrown straight up. Solving this equation for y, and substituting in v0y = v0 and vy = 0 m/s gives

2 y

vy

a

20= − (1)

To determine v0, we now apply the equation 2 2

0 2y y yv v a y= + to the situation where the ball is thrown upward at an angle of 52º relative to the horizontal. In this case we note that vy = v0 sin 52º, vy = 0 m/s, and y = y1 (the maximum height of 7.5 m reached by the ball). Solving for 2

0v , we find

( )2 12 20 1 0 2sin52 2 or

sin 52y

y ya y

v v a y v−2

= ° + =°

Substituting this expression for 2

0v into Equation (1) gives

12 1

2 27.5 msin 52 12 m

2 2 sin 52 sin 52

y

y y

a yv y

ya a

20

−2

°= − = − = = =° °

___________________________________________________________________________________________


73. REASONING The drawing shows the trajectory of the ball, along with its initial speed v0 and vertical displacement y. The angle that the initial velocity of the ball makes with the ground is 35.0°. The known data are shown in the following table:

y-Direction Data

y ay vy v0y t

+5.50 m −9.80 m/s2 +(46.0 m/s) sin 35.0° = +26.4 m/s ?

Since three of the kinematic variables are known, we will employ the appropriate equation

of kinematics to determine the time of flight. SOLUTION Equation 3.5b ( )21

20 y yy v t a t= + relates the time t to the three known

variables. The terms in this equation can be rearranged to as to place it in a standard quadratic form: 21

2 0 0y ya t v t y+ − = . The solution of this quadratic equation is

( )( )( )

( ) ( ) ( )( )( )

2 120 0

12

2 212

2

4

2

26.4 m/s 26.4 m/s 4 9.80 m/s 5.50 m0.217 s or 5.17 s

9.80 m/s

y y y

y

v v a yt

a

t

− ± − −=

− + ± + − − −= =

−

The first solution (t = 0.217 s) corresponds to the situation where the ball is moving upward

and has a displacement of y = +5.50 m. The second solution represents the later time when the ball is moving downward and its displacement is also y = +5.50 m (see the drawing). This is the solution we seek, so t = 5.17 s .

______________________________________________________________________________ 74. REASONING There are three velocities involved:

CRv = the initial velocity of the Car relative to the Road

CTv = the velocity of the Car relative to the first Truck

TRv = the velocity of the first Truck relative to the Road (speed = 11 m/s)

35.0°

v0

y

+x +y


Ordering these vectors by their subscripts in the manner discussed in Section 3.4 of the text, we see that vCR = vCT + vTR. The velocity TRv is known, and we will use the equations of kinematics to find vCT. The two trucks have the same velocity relative to the road, and thus have zero velocity relative to one another. Therefore, we will analyze the sports car’s jump across the 15-m gap as if both trucks were stationary. Neglecting air resistance, we treat the car as a projectile and find the initial speed v0 it must have, relative to the trucks, in order to reach the flat trailer. (This speed is, in fact, the speed vCT that we need to find.) We will use the relation

210 2x xx v t a t= +

(Equation 3.5a) to find the elapsed time. With ax = 0 m/s2, Equation 3.5a

becomes ( )2 210 02 0 m/sx xx v t t v t= + = . Thus the car is in the air for t seconds, where

0 0 cosx

x xtv v θ

= = (1)

In Equation (1), x = 15 m and θ = 16º, but v0 and t are not known. The heights of the trailer and the ramp are the same, so the car’s vertical displacement is zero. To use this fact, we turn to 21

0 2y yy v t a t= + (Equation 3.5b) and substitute both y = 0 m and

v0y = v0 sin θ:

00 yv t= 212 ya t+ 1 1

0 02 2 or or siny y yv a t v a tθ= − = − (2)

To eliminate the time, we substitute t from Equation (1) into Equation (2) and find that

210 0 02

0 sin or or

cos 2cos sin 2cos siny y

ya x a xxv a v v

vθ

θ θ θ θ θ− −⎛ ⎞

= − = =⎜ ⎟⎝ ⎠

(3)

SOLUTION Taking up as the positive direction, we use Equation (3) to calculate the speed vCT = v0 of the car relative to the trucks:

( )( )2

CT 0

9.80 m/s 15m17 m/s

2cos sin 2cos16 sin16ya xv vθ θ

− −−= = = =

° o

Since the ramp alters the direction of the car’s velocity but not its magnitude, the initial jump speed v0 is also the magnitude of the car’s velocity vCT relative to the truck before the car reaches the ramp: v0 = vCT. The sports car must overtake the truck at a speed of at least 17 m/s relative to the truck, so that the car’s minimum required speed vCR relative to the road is

CR CT TR 17 m/s 11 m/s 28 m/sv v v= + = + =


75. SSM REASONING The horizontal distance covered by stone 1 is equal to the distance

covered by stone 2 after it passes point P in the following diagram. Thus, the distance Δx between the points where the stones strike the ground is equal to x2, the horizontal distance covered by stone 2 when it reaches P. In the diagram, we assume up and to the right are positive.

θ

θ

θ

1

2

P

2x

Δ x SOLUTION If tP is the time required for stone 2 to reach P, then

2 0 P 0 P( cos )xx v t v tθ= = For the vertical motion of stone 2, 0 siny yv v a tθ= + . Solving for t gives

0 siny

y

v vt

a

θ−=

When stone 2 reaches P, 0 sinyv v θ= − , so the time required to reach P is

0P

2 sin

y

vt

aθ−

=

Then,

02 0 P 0

2 sin( cos )x

y

vx v t v

aθ

θ⎛ ⎞−⎜ ⎟= =⎜ ⎟⎝ ⎠


2 20

2 22 sin cos 2(13.0 m/s) sin 30.0 cos 30.0 14.9 m

–9.80 m/sy

vx

aθ θ− − ° °= = =

76. REASONING The three relative velocities in this situation are

vTG = the velocity of the Truck relative to the Ground (unknown) vTC = the velocity of the Truck relative to the Car (24.0 m/s, 52.0° north of east) vCG = the velocity of the Car relative to the Ground (16.0 m/s, due north)

The velocity of the truck relative to the ground may be expressed as the vector sum of the other two velocities: vTG = vTC + vCG . Note the fashion in which the “middle” subscripts on the right side of the equals sign are matched. See the diagram for an illustration of this vector sum. Because the vectors do not form a right triangle, we will utilize the component method of vector addition to determine the eastward and northward components of the resultant vector vTG. For convenience, we will take east as the +x direction and north as the +y direction. Once we know vTG,x and vTG,y, we will use the Pythagorean theorem to calculate the magnitude of the truck’s velocity relative to the ground, which, from the diagram, we expect to be larger than vTC = 24.0 m/s.

SOLUTION The vector vCG points due north, and thus has no x component. Therefore the x component of vTG is equal to the x component of vTC:

( )

TG, TC, CG, TC cos52.0 0 m/s

24.0 m/s cos52.0 14.8 m/s

x x xv v v v= + = +

= =

o

o

Next, we find the y component of the truck’s velocity relative to the ground. Noting that the vector vCG points due north, so that vCG,y = vCG, we have

( )

TG, TC, CG, TC CGsin 52.0

24.0 m/s sin 52.0 16.0 m/s 34.9 m/s

y y yv v v v v= + = +

= + =

o

o

The magnitude of vTG is found from the Pythagorean theorem:

East

North

vTG vCG

vTC

52.0°


( ) ( )2 22 2TG TG, TG, 14.8 m/s 34.9 m/s 37.9 m/sx yv v v= + = + =

77. REASONING Using the data given in the problem, we can find the maximum flight time t

of the ball using Equation 3.5b ( 1 20 2y yy v t a t= + ). Once the flight time is known, we can

use the definition of average velocity to find the minimum speed required to cover the distance x in that time.

SOLUTION Equation 3.5b is quadratic in t and can be solved for t using the quadratic

formula. According to Equation 3.5b, the maximum flight time is (with upward taken as the positive direction)

( ) ( )

( )( ) ( )

12 20 0 0 02

12

2 2

2

– – 4 – – 2

2

– 15.0 m/s sin 50.0 15.0 m/s sin 50.0 +2(–9.80 m/s ) (2.10 m)

–9.80 m/s

0.200 s and 2.145 s

y y y y y y

yy

v v a y v v a yt

aa

± ± += =

⎡ ⎤° ± °⎣ ⎦=

=

where the first root corresponds to the time required for the ball to reach a vertical

displacement of 2.10 my = + as it travels upward, and the second root corresponds to the time required for the ball to have a vertical displacement of 2.10 my = + as the ball travels upward and then downward. The desired flight time t is 2.145 s.

During the 2.145 s, the horizontal distance traveled by the ball is

[ ]0( cos ) (15.0 m/s) cos 50.0 (2.145 s) 20.68 mxx v t v tθ= = = ° =

Thus, the opponent must move 20.68 m –10.0 m 10.68 m= in 2.145 s – 0.30 s 1.845 s= . The opponent must, therefore, move with a minimum average speed of

vmin10.68 m1.845 s

5.79 m / s= =

78. REASONING The velocity OWv of the object relative to the water is the vector sum of the

velocity OSv of the object relative to the ship and the velocity SWv of the ship relative to the water, as indicated by Equation 3.7: OW OS SW= +v v v . The value of SWv is given in


the statement of the problem. We can find the value of OSv from the fact that we know the position of the object relative to the ship at two different times. The initial position is rOS1, and the final position is rOS2. Since the object moves with constant velocity,

OS OS2 OS1OS t t

Δ −= =

Δ Δr r r

v (1)

SOLUTION The first two rows of the following table give the east/west and north/south

components of the vectors OS2r and OS1−r . The third row of the table gives the components of OS OS2 OS1Δ = −r r r . Due east and due north have been taken as positive.

Vector

East/West Component

North/South Component OS2r

–(1120 m) cos 57.0°

= 26.10 10 m− ×

–(1120 m) sin 57.0°

= 29.39 10 m− ×

OS1−r

–(2310 m) cos 32.0°

= – 31.96 10 m×

+(2310 m) sin 32.0°

= 31.22 10 m× OS OS2 OS1Δ = −r r r

32.57 10 m− ×

22.81 10 m×

Now that the components of OSΔr are known, the

Pythagorean theorem can be used to find the magnitude.

ΔrOS2 2(–2.57 ) 2 ) 2.59= × + × = ×10 81 10 103 2 3 m m m( .

The direction of OSΔr is found from

φ = tan−1 2.81×102 m

2.57 ×103 m⎛⎝⎜

⎞⎠⎟= 6.24°

Therefore, from Equation (1),

vr r r

OSOS OS OS1 2.59 m / s, 6.24 north of west= =

−= × = °

ΔΔ Δt t

2 10 7 193 m

360 s.

Now that OSv is known, we can find OWv , as indicated by Equation 3.7: OW OS SW= +v v v .

The following table summarizes the vector addition:


Vector

East/West Component

North/South Component

OSv

–(7.19 m/s) cos 6.24° = –7.15 m/s

(7.19 m/s) sin 6.24° = 0.782 m/s

SWv

+4.20 m/s

0 m/s

OW OS SW= +v v v

–2.95 m/s

0.782 m/s

Now that the components of OWv are known, the

Pythagorean theorem can be used to find the magnitude. vOW

2 2(–2.95 m / s) m / s) 3.05 m / s= + =( .0 782

The direction of OWv is found from

φ = tan−1 0.782 m/s2.95 m/s

⎛⎝⎜

⎞⎠⎟ = 14.8° north of west

____________________________________________________________________________________________ 79. REASONING The drawing shows the

trajectory of the ball, along with its initial speed v0, horizontal displacement x, and vertical displacement y. The angle that the initial velocity of the ball makes with the horizontal is θ. The known data are shown in the tables below:

x-Direction Data

x ax vx v0x t

+26.9 m 0 m/s2 +(19.8 m/s) cos θ

y-Direction Data

y ay vy v0y t

+2.74 m −9.80 m/s2 +(19.8 m/s) sin θ

There are only two known variables in each table, so we cannot directly use the equations of

kinematics to find the angle θ. Our approach will be to first use the x direction data and obtain an expression for the time of flight t in terms of x and v0x. We will then enter this

θ

v0

y

x

+x +y


expression for t into the y-direction data table. The four variables in this table, y, ay, v0y, and t, can be related by using the appropriate equation of kinematics. This equation can then be solved for the angle θ.

SOLUTION Using the x-direction data, Equation 3.5a can be employed to find the time t

that the ball is in the air:

( )2 2120 0 since 0 m/sx x x xx v t a t v t a= + = =

Solving for t gives

( )0

26.9 m+ 19.8 m/s cosx

xtv θ

+= =

Using the expression above for the time t and the data in the y-direction data table, the

displacement in the y direction can be written with the aid of Equation 3.5b:

y = v0 y

t + 12 a

yt2

+2.74 m = 19.8 m/s( )sinθ⎡⎣

⎤⎦

+26.9 m

19.8 m/s( )cosθ

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= t

+ 12 −9.80 m/s2( ) +26.9 m

19.8 m/s( )cosθ

⎛

⎝⎜

⎞

⎠⎟

2

= t2

Evaluating the numerical factors and using the fact that sin θ /cos θ = tan θ , the equation

above becomes

( ) ( )2

9.04 m2.74 m 26.9 m tan

cosθ

θ−

+ = + +

Using 22

1 1 tancos

θθ= + , this equation can be rearranged and placed into a quadratic form:

( ) ( )29.04 m tan 26.9 m tan 11.8 m = 0θ θ− + −

The solutions to this quadratic equation are

( ) ( )( )( )

226.9 m 26.9 m 4 9.04 m 11.8 mtan 0.535 and 2.44

2 9.04 mθ

− ± − − −= =

−

The two angles are 1 1

1 2tan 0.535 28.1 and tan 2.44 67.7 θ θ− −= = ° = = ° ______________________________________________________________________________


80. REASONING Just after the ball is hit, its initial velocity has a magnitude v0 and components of v0x and v0y and is directed at an angle θ above the horizontal or x direction. The figure below shows the initial velocity vector and its components. As usual, we will use the Pythagorean theorem to relate v0 to v0x and v0y and will use trigonometry to determine θ.

SOLUTION The magnitude v0 of the initial velocity can be related to its components v0x and v0y by using the Pythagorean theorem, since the components are perpendicular to one another. This leads to following:

v

0= v

0x2 + v

0 y2 (1)

We can also use trigonometry to express the directional angle θ in terms of the components

v0x and v0y. Thus, we obtain: tanθ =v0yv0x

(2)

With no acceleration in the x direction, v0x remains unchanged throughout the motion of the ball. Thus, v0x must equal vx, the x component of the ball's final velocity. We have, then, that

v0x = vx = vcos28 . Substituting this result into equations (1) and (2) above gives the

following:

_____________________________________________________________________________

v0 = (vcos28)2 + (−vsin28)2 − 2ayy

= [(36 m/s)cos28]2 + [−(36 m/s)sin28]2 − 2(−9.80 m/s2 )(7.5 m) = 38 m/s

θ = tan−1 (−vsin28)2 − 2ayyvcos28

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= tan−1 [−(36 m/s)sin28]2 − 2(−9.80 m/s2 )(7.5 m)(36m/s)cos28

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= 33


81. SSM CONCEPTS (i) The y components of their velocities are equal: v0y = 10.0 m/s . Note that the initial velocity of the blue ball has no x component. (ii) The x components of their velocities are equal: v0x = 4.6 m/s . Note that the initial velocity of the yellow ball has no y component.

CALCULATIONS Based on the concept questions and answers, we know that the x and y

components of the red ball are v0x = 4.6 m/s and v0y = 10.0 m/s , respectively. The initial speed (launch speed) v0 is the magnitude of the initial velocity, and it, along with the directional angle θ, can be determined from the components:

_____________________________________________________________________________ 82. CONCEPTS (i) The final value vx of the horizontal component of the projectile’s velocity is

the same as the initial value v0x in the absence of air resistance. In other words, the horizontal motion occurs at a constant velocity of 25 m/s.

(ii) Yes. The time can be calculated as the horizontal distance (the range) divided by the

magnitude of the horizontal component of the projectile’s velocity. (iii) Yes. The value of the time calculated for the horizontal part of the motion is the same as

that for the vertical part, and can be used to analyze the vertical part of the motion. (iv) Since the projectile is launched from and returns to ground level, the vertical

displacement is zero.

CALCULATIONS From the constant-velocity horizontal motion, we find that the time is

For the vertical part of the motion, we know that the displacement is zero and that the acceleration due to gravity is −9.8 m/s2 , assuming that upward is the positive direction. Therefore, we can use Equation 3.5b to find the initial y component of the velocity:

v0 = v0x( )2 + v0y( )2= 4.6 m/s( )2 + 10.0 m/s( )2 = 11 m/s

θ = tan−1 v0y

v0x

⎛⎝⎜

⎞⎠⎟= tan−1 10.0 m/s

4.6 m/s⎛⎝⎜

⎞⎠⎟ = 65°

r = Rv0x

= 175 m25 m/s

= 7.0 s

y = v0yt + 12 ayt

2 or 0 m = v0yt + 12 ayt

2

v0y = − 12 ayt = − 1

2 −9.80 m/s2( ) 7.0 s( ) = 34 m/s


_____________________________________________________________________________

ANSWERS TO FOCUS ON CONCEPTS QUESTIONS

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