50

CHAPTER 4 MOTION IN TWO AND THREE DIMENSIONS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.

4-1. (a) Total displacement is (3.2 + 2.6) km 45° E of N + 4.5 km 50° W of N = 7.0 km , 5° E of N(see Problem 3, Chapter 3)(b) Avg vel = ∆r/∆t = 7.0 km/1.25 h, 5° E of N 5.6 km/h, 5° E of N=

(c) total distance 3.2 2.6 4.5 kmAvg speed 8.24 km/htime 1.25 h

+ += = =

4-2.11distance × × 1.50 × 10 mavg speed 29.9 km/s

time 1/ 2 yr 1/ 2 × (365.25 × 24 × 3600) srπ π= = = =

| Avg vel | = 11

7

2 × 2 × 1.50 × 10 m 19.0 km/s1/ 2 yr 1/ 2 × 3.16 × 10 s

rt

∆ = = =∆

r

4-3. 171 km/h = 47.5 m/sIf the bullet is fired at an angle θ, its position vector is given by rbul = 366 sin θ t i + 366 cos θ t jFor the bird, rbird = (47.5)t i + 30 jFor the bird to be hit rbul = r bird.Comparing the x-components, we have 366 sin θ = 47.5 θ = sin−1(47.5/366) = 7.45°Then d = 30 tan θ = 30 tan 7.45° = 3.93 m.Thus the hunter must aim 3.93 m ahead of the bird in order to hit the bird.

4-4. At t = 0, the velocity is 30 km/h N,=v or0 (30 km/h) .= +v i j At t = 10 s,30 km/h @ 45 W of N,= °v or

(21 km/h) (21 km/h) .= − +v i j At t = 20 s,

30 km/h W,=v or (30 km/h) 0 .= − +v i j At t = 30 s,30 km/h @ 45 S of W,= °v or

(21 km/h) (21 km/h) .= − −v i j At t = 40 s, the driver is

halfway around and v = 30 km/h S, or(30 km/h) 0 .= +v i j

4-5. r = (4 + 2t)i + (3 + 5t + 4t2)j + (2 − 2t − 3t 2)k(a) v = dr/dt 2 (5 8 ) (2 6 )t t= + + − +i j k(b) a = dv/dt 8 6= −j k

magnitude, | a | = 2 2 2(8) (6) 10 m/s+ =

direction θ = −tan−1 6 37° (37° below the -axis in the - plane)8

y z y⎛ ⎞ = −⎜ ⎟⎝ ⎠

t = 0

t = 10 s

t = 20 s

t = 30 s

t = 40 s

CHAPTER 4

51

4-6. (a) The components of v are vx = dx/dt and vy = dy/dt. Also, the components of a are ax = dvx/dtand ay = dvy/dt.

sinxv Ab bt= − cos yv Ab bt=2cos xa Ab bt= − 2sin ya Ab bt= −

(b) | v | = 2 2 2 2 2 2 ( sin ) ( cos ) (sin cosAb bt ab bt A b bt bt Ab− + = + =

| a | = 2 2 2 2( cos ) ( sin )Ab bt Ab bt− + − = Ab22 2 2

cos sinbt bt Ab+ =

†4-7. At t = 2.0 s, the missile has been falling with acceleration 20 (9.81 m/s )= −a i j and horizontalvelocity equal to the velocity of the airplane. This means the missile is still directly below the

airplane. Its displacement relative to the plane is 2 2 2(9.81 m/s )(2.0 s)0 0

2 2gt= − = − =r i j i j

0 (19.6 m) ,−i j or 19.6 m @ 90º below the direction of travel of the airplane. At t = 3.0 s, whichis 1.0 s after igniting the engine, the acceleration is 2 2(6.0 m/s ) (9.81 m/s ) .= −a i j Its

displacement during this 1.0 s interval is 2 2

1 2 2xa t gt= −r i j

2 2 2 2 2 2 2(6 m/s ) (6 m/s )(1.0 s) (9.81 m/s )(1.0 s)2 2 2 2

t gt= − = − =i j i j (3.0 m) (4.9 m) .−i j Now

the missile’s total displacement relative to the plane is2 (0 m) (19.6 m) (3.0 m) (4.9 m)= − + −r i j i j (3.0 m) (24.5 m) .= −i j The magnitude is

2 22 (3.0 m) (24.5 m) 24.7 m.r = + = The direction is given by 1 24.5tan 83.0 ,

3θ − −= = − ° or

24.7 m @ 83º below the direction of travel of the plane.

4-8. 2 2 3(5 4 ) (3 2 ) 0 m.t t t t= + + + +r i j k 2(5 8 ) (6 6 ) 0 m/s.d t t tdt

= = + + + +rv i j k

22

2 (8) (6 12 ) 0 m/s .d d tdt dt

= = = + + +r va i j k At t = 2.0 s, (21 m/s) (36 m/s) 0 m/s,= + +v i j k

which gives a speed of 2 2(21 m/s) (36 m/s)v = + = 42 m/s.

†4-9. ,average t∆=∆

rv where ∆r is the total displacement and ∆t = 1.5 h is the total elapsed time. To find

the total displacement, find the displacement during each part of the trip:1 (300 km/h @ 30 N of E) × 0.50 h 150 km @ 30 N of E.∆ = ° = °r Taking the y direction to

point N and the x direction to point E, this is 1 (150 km)( cos 30°) (150 km)( sin 30°)∆ = +r i j(130 km) (75 km) .= +i j For the second part of the trip,

2 (300 km/h @ 30 W of S) × 1.0 h∆ = °r 300 km @ 30 W of S.= ° In terms of x and y, thisis 2 (300 km)( sin 30°) (300 km)( cos 30°) (150 km) (260 km) .∆ = − − = − −r i j i j The total

displacement is 1 2 20 km 185 km .∆ = ∆ + ∆ = − −r r r i j The average velocity is average t∆=∆

rv

CHAPTER 4

52

20 km 185 km 13.3 km/h 123 km/h .1.5 h

− −= = − −i j i j To give this as a

speed and heading, use 2 213.3 123 km/h 124 km/h.averagev = + = Since

both velocity components are negative, the vector is located in the thirdquadrant (between W and S), so we can give the direction

as 1 123tan 83.8 S of W.13.3

θ −= = ° Thus vaverage = 124 km/h @ 83.8º S of

W.

The average acceleration is 2 1 ,average t t−∆= =

∆ ∆v vva where v1 is the velocity during the 0.5, v2 is

the velocity during the next 1.0 h, and ∆t is the time interval during which the velocity changes.Since no value is given for ∆t, we can calculate ∆v and give the answer symbolically. Using thedefinitions above, 1 (300 km/h) cos 30 (300 km/h) sin 30= ° + °v i j (260 km/h) (150 km/h) .= +i j30° W of S is the same as 240° N of E, so 2 (300 km/h) cos 240 (300 km/h) sin 240= ° + °v i j

(150 km/h) (240 km/h) .= − −i j Then (410 km/h) (410 km/h) ,∆ = − −v i j

and (410 km/h) (410 km/h) .average t− −=

∆i ja As a magnitude and heading, this

is 410 2 km/h @ 45 S of W,average t= °

∆a or 580 km/h @ 45 S of W.average t

= °∆

a Even though we

don’t have a numerical value, it is important to note that the average acceleration is not zero, eventhough the speed of the airplane is constant. The acceleration is caused by the change in directionof the velocity.

4-10. cos .A bt Bt= +r i j sin .d bA bt Bdt

= = − +rv i j2

22 cos 0 .d d b A bt

dt dt= = = − +r va i j The equation

for the speed is 2 2 2 2( sin ) .x yv v v bA bt B= + = +

†4-11. 90 (500 15 )t t= + −r i j m. 90 15 m/s.ddt

= = −rv i j The equation

for the speed is2 2 2 2(90 m/s) (15 m/s) 91 m/s.x yv v v= + = + = The direction

of the velocity is 1 1 15tan tan 9.5°.90

y

x

vv

θ − − −⎛ ⎞= = = −⎜ ⎟⎝ ⎠

This is 9.5° below the x axis.

vaverage

θE

N

90 m/s

θ

y

x15 m/s

CHAPTER 4

53

4-12. Measure time from when the ball is thrown. At that instant, thereceiver has traveled (7 m/s)(2.0 s) = 14 m along the x directionrelative to the quarterback. The receiver is at an initialdisplacement 0 (14 m) (20 m)= +r i j relative to the quarterback.Since he’s running at a constant velocity of vR = 7.0i m/s, his totaldisplacement is (14 7 ) 20 R t= + +r i j m. Once it’s thrown, thedisplacement of the ball relative to the quarterback is

, , ( sin ) ( cos ) ,B B x B y B Bv t v t v t v tθ θ= + = +r i j i j where vB = 15 m/s.

Mathematically, the ball is caught

when B R=r r : 1sin cos .2

θ θ= = Since the

x components must be equal and the y components must be equal, we get two simultaneousequations:(15sin ) 14 7(15cos ) 20

t tt

θθ

= +=

Use the second equation to get 20 4 .15cos 3cos

tθ θ

= = Substitute this into the first equation to get

4 415sin 14 7 .3cos 3cos

θθ θ

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Simplifying gives 30sin 21cos 14.θ θ− =

This transcendental equation can be solved by trial and error, by graphical means, or by use oftools like Solver in Excel or a Solve Block in MathCAD. The result is θ = 57.6°. The flight time

for the ball is 4 2.5 s.3cos 57.6

t = =°

†4-13. 3 2 (3 2 ) .d d dtdt

= = + ⇒ = +va i j v i j Integrate to find v: 0 0

(3 2 ) ,t

d dt= +∫ ∫v

vv i j which gives

0 3 2 .t t− = +v v i j The problem states that both components of velocity are initially zero, so we

get 3 2 m/s.t t= +v i j Since ,ddt

= rv we can follow the same procedure that was used to get v

from a: 0

22

00

3(3 2 ) (3 2 ) .2

t td t t dt d t t dt t= + ⇒ = + ⇒ − = +∫ ∫r

rr i j r i j r r i j The problem says

the particle starts moving at the origin, so both components of r0 are zero. 2

23 m.2t t= +r i j

4-14. 162.3 km/h = 45.1 m/s

Time to travel 20m: 20 0.44 s4501

t = =

The distance fallen vertically: 12

z = gt2 = 21 (9.8)(0.44) 0.95 m2

=

x

y

vR

vBθ

CHAPTER 4

54

†4-15. Assume the divers launch themselves horizontally from the edge of the cliff, so 0.oyv = The time

to fall a distance h = 64 m is 2

2 2(36 m) 2.71 s.9.81 m/s

htg

= = = During this time, the diver must

travel a horizontal distance of at least x = 6.4 m, so the minimum horizontal velocity required

is 06.4 m 2.4 m/s.2.71 sx

xvt

= = =

4-16. From Example 6, t = 3.19 s (An extra digit has been kept so the final results of this problem cansafely be rounded to the two digits required by the data given in the Example.)(a) 83.3 m/sxv = (83 m/s to two digits) (ax = 0, so the horizontal velocity is constant)

2(9.81 m/s )(3.19 s) 31.3 m/s.yv gt= − = − = − To two digits, vy = −31 m/s.

(b) 2 2 2 2(83.3 m/s) (31.3 m/s) 89 m/s.x yv v v= + = + = The speed has increased by about 6

m/s.†4-17. The stunt car will fall 2 m while it travels 24 m horizontally.

The time to fall 2 m is 2

2 2(2 m) 0.639 s.9.81 m/s

htg

= = = The

car must travel 24 m horizontally in 0.639 s, so its horizontalspeed must be 38 m/s. (This is about 135 kph, or 84 mph.)

4-18. v = v0 + at = (3i + 2j) + (i − 4j)t = (3 + t) i + (2 − 4t) jThe maximum y-coordinate is reached when the y-component of the velocity is zero. Thus, 2 − 4t= 0 or 0.5 st =

r = 0or + v0t + 1

2a t2 = 3t i + 2t j + 0.5t2 i − 2t2 j = (3t + 0.5t 2) i + (2t − 2t2) j.

At t = 0.5 s, = (1.625 + 0.5 ) mr i j

4-19. The maximum distance, xmax = 20v /g

(i) v0 = maxx g = 441 × 9.8 65.7 m/s=

(ii) v0 = 889 × 9.8 93.3 m/s (336 km/h)=

4-20. xmax = 20v sin 2θ/g with θ = 12°, xmax = 250 m

20v = 2 2 2 2max 9.81 × 250 m /s 6030 m /s

sin 2 sin 24gx

θ⎛ ⎞= =⎜ ⎟°⎝ ⎠

o 77.7 m/sv =

max ht2 2 20

maxsin 6030 sin 12 m 13.3 m2 2 × 9.81

vzg

θ °= = = =

4-21. (a) The horizontal range is given by202 sin cos .target

vxgθ θ= Using sin 2 2sin cosθ θ θ= gives

20

sin 2 .targetgxv

θ = Thus 120

1 sin ,2

targetgxv

θ − ⎛ ⎞= ⎜ ⎟

⎝ ⎠ so

21

2

1 (9.81 m/s )(12 500 m)sin = 7.25°.2 ( m/s)

,θ − ⎡ ⎤= ⎢ ⎥700⎣ ⎦

24 m

v0

2 m

CHAPTER 4

55

(b) In this part of the problem, it will be necessary to assume that all the digits given aresignificant, and that the values of g are specified exactly as given. Suppose the gunner uses g =9.80 m/s2 in the calculations. Then the elevation angle would be calculated using

20

2

sin 2 * ,9 80 m/stargetvx θ=

. where xtarget is the given distance to the target and θ * is the wrong angle

(associated with g = 9.80 m/s2). As in part (a), sin 2θ * = 2

20

(9.80 m/s ).

vt argetx

But the actual range

will be 20

2

sin 2 * .9 81 m/smaxvx θ=

. Substituting the expression for sin 2θ * gives 9.80

9.81max targetx x ⎛ ⎞= ⎜ ⎟⎝ ⎠

9.80(12,500 m) 12,487 m.9.81

⎛ ⎞= =⎜ ⎟⎝ ⎠

The gunner will miss the target by 13 m.

4-22. Initial speed, 0 max 1500 × 9.8 121.2 m/sv x g= = =2 20 max

maxsin 375 m2 4

v xzg

θ= = =

0flight

2 sin 2 × 121.2 17.5 s2 × 9.8

vtg

θ= = =

4-23. (a) vty = 0 = viy − gt

t = iyvg

= 387 × 10 sin 45 1.74 s

60 × 60 9.8⎛ ⎞

=⎜ ⎟⎝ ⎠

(b) zmax = 2

2iyvg

= 287 sin 45

3.6⎛ ⎞⎜ ⎟⎝ ⎠

1 14.9 m2 × 9.8

=

(c)22

0max

87 1 59.6 m 3.6 9.8

vxg

⎛ ⎞= = =⎜ ⎟⎝ ⎠

4-24. Denote Berlin by x1, g1 and Buenos Aires by x2, g2.2 20 0

1 21 2

sin 2 sin 2v vx xg g

θ θ= =

Then 1

2

xx

= 2

1

gg

≥ x2 = 1 1

2

x gg

= 68.11 m × 9.8128 69.22 m9.7967

=

0.11 mx∆ =

4-25. 6 3 0 max s 10 × 9.8 3.13 × 10 m/sv x= = =

zmax = 2y

2vg

= 3 2 2

5(3.13 × 10 ) sin 45 2.5 × 10 m2 × 9.8

=

tflight = 3

02 sin 45° 2 × 3.13 × 10 sin 45 452 s9.8

vg

°= =

CHAPTER 4

56

4-26. (a) xmax = 20v /g ≥ v0 = maxx g = 50 × 9.8 m/s 22 m/s=

(b) The circumference of the circle is 2π × 1.0 m = 6.28 m.

trev = distspeed

= 6.28 m22 m/s

= 0.285 s

Thus, number of rev/s = rev

1t

= 10.285

s 3.5 rev/s=

4-27. xmax = 20 sin 2v

gθ =

2 (26) sin 70° 64.8 m9.8

=

tflight = 0 2 sin vg

θ = 2 (26) sin 35 3.04 s9.8

° =

The amount of water in the air at any given instant is 280 × 3.04 14.2 liters60

=

4-28. xmax = 20v /g ≥ v0 = max gx = 730 × 9.8 m/s 85 m/s=

tflight = 02 sin vg

θ = 2 × 85 × sin 45 s 12 s9.8

° =

4-29.2 2 2 2 2 20 0 0 02 sin cos sin 2 sin cos sin, . .

2 2max max max maxv v v vx y x y

g g g gθ θ θ θ θ θ= = = ⇒ = This

gives sin2cos ,2

θθ = or tan 4θ = ⇒ θ = 76°.

4-30. 20 2 2(9.81 m/s )(0.60 m) 3.43 m/s.y maxv gy= = = 0

2

2 2(3.43 m/s) 0.70 s.9.81 m/s

yflight

vt

g= = = (The

exact angle at which the balls are thrown is not relevant.)

†4-31. 2.0 4.5 (2.0 4.5 ) .d d dtdt

= = − ⇒ = −va i j v i j Integrate to find v following the method used in

Problem 4-13: 0 0(2.0 4.5 ) 2.0 4.5 .

tdt t t− = − = −∫v v i j i j The problem states

that 0 10 25 ,= − +v i j so we get 2.0 4.5 ( 10 25) (2.0 10) (25 4.5 )t t t t= − + − + = − + −v i j i j i jm/s.At t = 3.0 s, the velocity is = (6.0 10) (25 13.5) 4.0 11.5 m/s.− + − = − +v i j i j The speed is

2 2 2 2(4.0 m/s) (11.5 m/s)x yv v v= + = + = 12 m/s.

Since ,ddt

= rv we can follow the same procedure that was used to get v from a:

0 0[(2.0 10) (25 4.5 ) ] [(2.0 10) (25 4.5 ) ]

td t t dt d t t dt= − + − ⇒ = − + − ⇒∫ ∫

r

rr i j r i j

2 20 ( 10 ) (25 2.25 )t t t t− = − + −r r i j m. The problem says the particle starts moving at the origin,

so both components of r0 are zero. 2 2( 10 ) (25 2.25 ) m.t t t t= − + −r i j At t = 3.0 s,21 55 m.= − +r i j

CHAPTER 4

57

4-32. 00

2 sin .2sin

flightflight

gtvt vg

θθ

= ⇒ = 2 22

02 sin cos 2sin cos2sin 2 tan

flight flightmax

gt gtvxg gθ θ θ θ

θ θ⎛ ⎞

= = =⎜ ⎟⎝ ⎠

⇒

21tan .

2flight

max

gtx

θ − ⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠

2 21 (9.81 m/s )(6.0 s)tan

2(75 m)θ − ⎡ ⎤

= =⎢ ⎥⎣ ⎦

67°.

4-33.2 20

02 sin cos (9.81 m/s )(15 m)

2sin cos 2sin10 cos10max

maxv gxx v

gθ θ

θ θ= ⇒ = = =

° ° 21 m/s.

4-34. The time to fall a distance h is 2 .htg

= The object must also

travel the same horizontal distance h during this time,

so 0 .2

h ghvt

= = After time t the vertical velocity will

be 2 ,yv gt gh= − = − so the final velocity at time of impact is

2 .2gh gh= −v i j

†4-35. The time to reach maximum height is 2.25 s 1.125 s.2heightt = = 0 y

height

vt

g= ⇒

20 (9.81 m/s )(1.125 s) 11.0 m/s.y heightv gt= = = If we say the launch point was at y = 0, then the

equation for the position below the launch point is 2

0 2ygty v t= − with 0 11.0 m/s.yv = This

gives 2 2(9.81 m/s )(4.00 s)(11.0 m/s)(4.00 s) 34.5 m.2

y = − = − The cliff is 34.5 m high. Since

the launch angle was 45°, 0 0 11.0 m/s.x yv v= = The total horizontal distance traveled since

launch is 0 (11.0 m/s)(2.25 s 4.00 s) 68.8 m.x totalx v t= = + =4-36. Choose the origin of the coordinate system to be at the bottom of

the cliff. Then 2

0 0 .2y

gty y v t= + − 0 030 m; 25 m/s.y h v= = =

0 0 sin 30 12.5 m/s.yv v= ° = At the bottom, y = 0, so we must

solve this quadratic equation: 20 30 12.5 4.91 .t t= + − The rootsare given

by212.5 (12.5) 4(4.91)(30)

1.51 s, 4.06 s.9.81

t− ± +

= = −−

The

final result must be positive, so t = 4.06 s.

4-37. The flight time to the goal is 0

25 m 0.391 s.cos (65 m/s) cos10xt

v θ= = =

° The height of the puck at

this time is 2 2 2

0(9.81 m/s )(0.391 s)sin (65 m/s)(sin10 )(0.391 s) 3.66 m.

2 2gty v tθ= − = ° − =

The puck will pass 2.2 m above the goal.

v0

h

h

v0

h

30°

CHAPTER 4

58

4-38. (a)202 sin cos

maxvx

gθ θ= ⇒

2

0(9.81 m/s )(240 m)

2sin cos 2(sin14 )(cos14 )maxgxv

θ θ= = =

° ° 70.8 m/s.

(b)2

2

2(71.4 m/s) sin14 cos149.81 m/smaxx ° °= = 244 m. The ball will overshoot by about 4 m.

(c)2

2

2(70.8 m/s) sin14.5 cos14.59.81 m/smaxx ° °= = 248 m. The ball will overshoot by about 8 m.

4-39.2 2 20 02 sin cos sin, .

2max maxv vx y

g gθ θ θ= = For θ = 45°, 1sin cos .

2θ θ= = Substituting in the

equations for xmax and ymax gives2 2 20 0 02 , .

2 4max maxv v vx yg g g

= = = Taking the ratio gives 1,4

max

max

yx

=

which is the desired result.

4-40.2 20

max sin =2

vzg

θ = 8.667 ft (i)

20

max sin 2 = vx

gθ = 37 ft (ii)

Dividing (i) by (ii) gives 2sin 0.2342

sin 2θθ

=

2 2sin sin 1= = tan = 0.2342sin 2 4 sin cos 4

θ θ θθ θ θ

θ = tan−1 (4 × 0.2342 ) 43.1= °From (ii) 2

0v = 37 × 32.2/sin 86.2° = 1194 ft2/s2

0 34.6 ft/sv =

†4-41.2 2 20 02 sin cos sin, .

2max maxv vx y

g gθ θ θ= = Setting these equal to each other gives

2 2 20 02 sin cos sin ,

2v v

g gθ θ θ= from which we get sin tan 2,

cosθ θθ

= = or 1tan 2θ −= = 63.4°.

4-42. tflight 02 sin 2 × 38 sin 52 6.1 s

9.8v

gθ °= = =

In order to catch the ball, the fielder must run with a minimum speed of 45/6.1 = 7.4 m/s.Therefore with 8 m/s the fielder will be able to catch the ball.

†4-43. A rough sketch (not to scale) is shown. The coordinates ofthe projectile are

2

0

0

(sin )2

(cos )

gty h v t

x v t

θ

θ

= + −

=The coordinates of the target are (xmax,0). The flight time

then is0

.cosmaxxt

v θ= Set y = 0 and substitute this time.

v0

h

θ

xmax

y

x

CHAPTER 4

59

After some rearranging, the equation becomes 2

02 20 0

2 sin 2 0.cos cos

max maxx xg v hv v

θθ θ

⎛ ⎞ ⎛ ⎞− − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ We

can convert this to a quadratic equation with tan θ as the variable by using the following

trigonometric steps: sintan .cos

θθθ

= 2 2

22 2 2

sin 1 cos 1tan 1.cos cos cos

θ θθθ θ θ

−= = = − Thus

22

1 1 tan ,cos

θθ

= + and we have 2

220

(1 tan ) 2 tan 2 0.maxmax

gx x hv

θ θ⎛ ⎞

+ − − =⎜ ⎟⎝ ⎠

Multiplying by

202max

vgx

⎛ ⎞⎜ ⎟⎝ ⎠

and rearranging finally gives this equation:

2 22 0 0

2 2tan 2 tan 1 2 0.maxmax max

v vx hgx gx

θ θ⎛ ⎞ ⎛ ⎞

− + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Using xmax = 12 × 103 m, v0 = 600 m/s, and g =

9.81 m/s2 gives 202max

vgx

⎛ ⎞⎜ ⎟⎝ ⎠

= 2.548 × 10-4 m-1. Substituting h = 50 m and doing the rest of the

arithmetic finally gives 2tan 6.116 tan 0.9745 0.θ θ− + = The solution is26.116 6.116 4(0.9745)

tan 5.953, 0.1635.2

θ ± −= = The two angles are the inverse tangents

of these: θ = 80.5°, 9.29°. It turns out that most projectile aiming problems like this have twosolutions. The difference between the two angles is that the larger elevation will give a longerflight time and the projectile will strike the target at a larger angle than for the smaller elevation.

4-44. (a)2

0, .2

gty x v t= − = The distance d down the slope is related

to x and y by the slope angle of 45°: cos 45 , sin 45 .x d y d= ° = °Substitute these into the equations for x and y. Solve the x

equation for t: 0

cos 45 ,dtv

°= Substitute into the y equation:

2

0

cos 45sin 45 .2g dd

v⎛ ⎞°° = ⎜ ⎟⎝ ⎠

Solve for d:

20

2 2

2 sin 45 2 sin 45°. = 2,cos 45 cos 45°

vdg

°=°

so 202 .vd

g= v0 = 110 km/h = 30.6 m/s

2

2

2(30.6 m/s) 135 m.9.81 m/s

d⇒ = =

(b) Skiers are able to land farther down the slope by arching their bodies forward over the skis.This moves their center of gravity forward, reduces aerodynamic drag, and actually providessome aerodynamic lift which carries them farther down the slope than would be predicted byprojectile motion analysis.

4-45. x = 90 = (70 cos θ) t (i)

y = 0 = (70 sin θ) t − 12

gt2 (ii)

Substitute for t from (ii) in (i) to get x = 20 sin 2v

gθ

v0

dy

x45˚

CHAPTER 4

60

θ = 12

sin−1 20

x gv

⎛ ⎞⎜ ⎟⎝ ⎠

= 12

sin−1 2

90 × 9.8(70)

⎛ ⎞⎜ ⎟⎝ ⎠

= 5.18°

when the arrow is misaimed by 0.03° in the vertical direction,

y = x tan θ − 12

g

29070

⎛ ⎞⎜ ⎟⎝ ⎠

2

2 2

1 1 9 190 tan (5.21 ) 9.8cos 2 7 cos 5.21θ

⎛ ⎞= ° − ⎜ ⎟ °⎝ ⎠= −3.9 × 10−2 m 3.9 cm (Arrow hits bull’s-eye)= −

A misaim of 0.03° in the horizontaldirection will give the arrow acomponent of velocity in thez-directionv0z = v0 cos 5.18° sin 0.03The deviation in the horizontaldirection from the center of thebull’s-eye is

vozt = (70 cos 5.18° sin 0.03°) 9070 cos 5.18°

= 4.7 × 10−2 m 4.7 cm=Therefore the arrow will hit the bull’s-eye.

4-46. (a) xmax = 20 sin 2v

gθ ⇒ sin 2θ = 2

0

maxx gv

sin 2 θ = 2 2(700 m × 9.8 m/s ) / (630 m/s) = 0.01728

θ = 12

sin−1 (0.01728) = 0.50°

If d is the distance above the target, then tan θ = d/700 or d = 700 m tan 0.5°6 md =

(b) zmax = 2 20 sin

2v

gθ =

2 2630 sin (0.5) 1.5 m2 × 9.8

=

(c) tflight = 02 sin vg

θ = 2 × 630 × sin (0.5) 1.1 s9.8

=

4-47. xmax = 20v sin 2θ/g. Let the target distance be x

x − 180 = 20v sin 14.667°/g (i)

x + 120 = 20v sin 15.167°/g (ii)

Dividing (i) by (ii) gives180120

xx

−+

= sin 14.667sin 15.167

°°

⇒ (x − 180) sin 15.167° = (x + 120) sin 14.667°

x = 77.480.00844

= 9180 m

CHAPTER 4

61

From (i) 9180 m = 180 m = 20v sin 14.667°/g

20v = 3.483 × 105 m2/s2

sin 2θ = 20

xgv

= 5

9180 × 9.83.48 × 10

= 0.23

θ = 1/2 sin−1 0.23 7.49 7 29 '= ° = °

4-48.2 2 20 02 sin cos sin, .

2max maxv vx y

g gθ θ θ= = Dividing the second equation by the first gives

2sin tan ,4sin cos 4

max

max

yx

θ θθ θ

= = from which we get 1 14 4(2.5)tan tan 63.4 .5

max

max

yx

θ − −⎛ ⎞ ⎡ ⎤= = = °⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠2

0

2 2(9.81 m/s )(2.5 m)7.8 m/s.

sin sin 63.4maxgy

vθ

= = =°

†4-49. Consider the motion of two of the projectiles. Call their positions (x1, y1) and (x2, y2) and assumethat they were launched with elevation angles θ1 and θ2, respectively. The positions are given by

2

1 0 1 1 0 1cos , sin2

gtx v t y v tθ θ= = − and 2

2 0 2 2 0 2cos , sin .2

gtx v t y v tθ θ= = − For the projectiles

to collide, they must have the same positions at the same time: x1 = x2 and y1 = y2. This gives the

following pair of equations: 2 2

0 1 0 2 0 1 0 2cos cos , sin sin .2 2

gt gtv t v t v t v tθ θ θ θ= − = − Canceling

terms gives 1 2 1 2cos cos , sin sin .θ θ θ θ= = Dividing the second equation by the first gives

1 2tan tanθ θ= as the requirement for the projectiles to collide. Since we assumed that theprojectiles were launched at different angles, we arrive at a contradiction, and we conclude thatthe individual projectiles will never collide during their flight.

4-50. Neglecting air resistance, the net force on the projectile is always − mgj (taking the +y directionto point up). If the motion is instantaneously uniform circular motion at the top of the trajectory,then the magnitude of the centripetal force must be mg, and the centripetal acceleration must be g.

Thus20 ,xv gr

= which gives 20xvrg

= for the radius of the osculating circle.

†4-51. Take the x and z directions to point as shown. (The y direction is up,out of the page.) The projectile has two horizontal components ofvelocity: 0 cos ,horizontal Sv vθ= +v i k where v0 is the muzzle speed ofthe projectile, θ is the elevation angle of the projectile, and vS is thespeed of the ship. Neglecting air resistance, the flight time for the

projectile is 02

2 sin 2(720 m/s)(sin 30 ) 73.5 s.9.81 m/sflight

vtg

θ °= = = The

speed of the ship is 45 km/h = 12.5 m/s, so the net horizontaldisplacement of the projectile is horizontal flightt=r v

3[(720 m/s)(cos 30 ) (12.5 m/s) ](73.5 s) (45.8 × 10 m) (918 m) .= ° + = +i k i k The total horizontal

range is 2 2 45.8 km.x zr r r= + = The change in the total range due to the ship’s motion isnegligible to three significant digits; however, failure to compensate for the ship’s forward speedwould cause the projectile to miss its target by nearly 1 km.

x

z

CHAPTER 4

62

4-52. x = v0t cos θ (i) z = v0t sin θ − 12

g t2 (ii)

t = 0 cos

xv θ

[from (i)] (iii). Substituting (iii) into (ii) gives z = x tan θ − 2

2 2 0

12 cos

gxv θ

13 = 50 tan θ − 2

2

1 (9.8)(50)2 25

sec2 θ.

13 = 50 tan θ − 19.6(1 + tan2 θ)19.6 tan2 θ − 50 tan θ + 32.6 = 0This equation has imaginary roots, so stone cannot hit the window.

z = 50 tan θ − 19.6(1 + tan2 θ)Let y = tan θ, z = −19.6y2 + 50y = 19.6

and 39.2 50dz dz dy dz yd dy d dyθ θ

⎛ ⎞ ⎛ ⎞= = − +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

At maximum, dz/dθ = 0, so that dz/dy = 0 = −39.2y + 50y = 50/39.2 = 1.28 = tan θz = 50(1.28) − 19.6(1 + 1.282) 12.3 m=

†4-53. From the diagramv0x = v0 cos θv0y = v0 sin θ (i)In the xy-framex = v0x t (ii)

y = v0y t − 12

gt2 (iii)

Alsoy = l sin αx = l cos α (iv)When the projectile hits the incline

y = l sin α = v0yt − 12

gt2 (v)

from (ii) t = x/v0x = l cos α/v0x (vi)

l

Substitute t from (vi) in equation (v) to get the range as measured along the incline.

l = 2 2

0 2 2

0 0

1 cos α 1 cos αsin sin α cos 2 cos

l lv gv v

θθ θ

⎧ ⎫−⎨ ⎬

⎩ ⎭i

Solving for l, gives

l = { }2

202 cos sin tan α cos cos α

vg

θ θ θ−

( )2 2

02 cos tan tan α cos α

vlg

θ θ= −

l will be maximum when dldα

= 0

( )202 cos 2 + (tan α) 2 cos sin = 0

cos αvdl

d gθ θ θ

θ=

CHAPTER 4

63

or cos 2θ = − tan α sin 2θ, cot 2θ = − tan α = cot (α ± π/2)

Thus 1 (α / 2)2

θ π= ±

4-54. This problem is almost identical to 4-12. Follow the procedure given there. Using the slightlydifferent numerical data for this problem leads to these equations for the position of the receiverand the position of the ball relative to the quarterback after the ball is thrown:

(16 8 ) 15 ,R t= + +r i j (20sin ) (20 cos ) .B t tθ θ= +r i j The condition B R=r r gives this pair ofequations: (20sin ) 16 8 ; (20 cos ) 15.t t tθ θ= + = Use the second equation to

get 15 3 .20cos 4 cos

tθ θ

= = Substitute this into the first equation to get

4 420sin 16 8 .3cos 3cos

θθ θ

⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Simplifying gives 15sin 16 cos 6.θ θ− = Solving the

transcendental equation gives 62.7°, or θ = 63° to two digits. The flight time for the ball is3 1.64 s.

4cos 62.7t = =

°Call the elevation angle α. To find this angle, use the fact that we know both the range and theflight time. In terms of the initial speed of the ball and the angle α, we have

20 02 sin cos 2 sin; .v vr t

g gα α α= = Solve the second equation for v0 and substitute that

expression into the first equation: 2 2 2sin cos cos2 .

2sin 2sin 2 tangt gt gtr

gα α α

α α α⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠ The

range can also be written in terms of vB and t: ,Br v t= so we have 2

,2 tanB

gtv tα

= from which we

get 2

1 1 (9.81 m/s )(1.64 s)tan tan 21.9 ,2 2(20 m/s)B

gtv

α − −⎛ ⎞ ⎡ ⎤= = = °⎜ ⎟ ⎢ ⎥

⎣ ⎦⎝ ⎠ or α = 22°.

†4-55. The total instantaneous velocity of a point on the rim of thetire is the vector sum of the forward velocity of the tractorand the tangential velocity of the point, .tot tan= +v v u Thistotal instantaneous velocity must be zero when the point ison the road (θ = 270°), or the tire will be slipping. Thus thetangential speed of a point on the rim must be equal to theforward speed u of the tractor.

In the reference frame of the tractor, the horizontalcomponent of velocity of a glob is always zero and thevertical component of the velocity is cos ,yv u θ= which becomes the y component of the initial

velocity v0y of a launched glob. The maximum height above the launch point is2 2 20 cos .

2 2y

max

v uyg g

θ= = To find the maximum height above the ground, we must add the height

of the launch point above the ground, which is sinR R θ+ as shown in the diagram. So the

height above the ground is 2 2cos(1 sin ) .

2uh R

gθθ= + +

θ

u

θu cosθ

R

R sinθ

CHAPTER 4

64

The maximum height depends on θ. To find the maximum possible height reached by aglob, differentiate h with respect to θ and set the derivative equal to zero:

2 cos sincos 0,dh uRd g

θ θθθ

= − = from which we get 2sin .maxgRu

θ = For R = 0.80 m and u = 30

km/h = 8.33 m/s, this gives θmax = 6.49°, from which we get2 2

2

(8.33 m/s) (cos 6.49 )(0.80 m)(1 sin 6.49 ) 4.4 m.2(9.81 m/s )

h °= + ° + =

4-56. 40 km/h = 11.11 m/s(a) The equation for the projectile is

(700 cos θ)t i + 21(700 sin )2

t gtθ⎡ ⎤−⎢ ⎥⎣ ⎦j = rproj

The equation for the ship is (1500 − 11.11t) i = rship

When the projectile hits the ship, the coefficient of j = 0, so

(700 sin θ) t − 12

gt 2 = 0 (i)

Also, the coefficients of i must be equal, so(700 cos θ) t = 15,000 − 11.11t

t = 15,000700 cos 11.11θ +

Substituting this into (i) gives

700 sin θ = 12

gt = 12 g 15, 000

(700 cos 11.11)θ +sin θ (700 cos θ + 11.11) = 105An iterated solution gives 8.6θ ≈ °

(b) The time interval is t = 15,000 21.3s(700 cos 8.6° 11.11)

=+

4-57. If the ship does not move, the time of flight is given by tflight = 2v0 sin θ/gθ is given by xmax = 17,000 m = 2

0v sin 2θ/g = 7002 sin 2θ/9.81

sin 2θ = 2

9.81 × 17, 000700

= 0.34

θ = 12

sin−1 (0.34) = 9.94°

tflight = 2 × 700 m × sin 9.949.81

°⎛ ⎞⎜ ⎟⎝ ⎠

s = 24.67 s

In this time the ship would have moved 8.33 m/s × 24.67 s = 205.5 m.Thus the new xmax (by the Pythagorean theorem) is

2 217,000 205.4+ = 17,001 m

xmax = 17001 m = 20v sin 2θ/g ≥ θ = 1

2 sin−1 2

17001 × 9.8700

⎛ ⎞⎜ ⎟⎝ ⎠

= 9.94°

so the time of the flight is still the same, so the projectile will arrive there when the ship does.9.94 9 56 'θ = ° = °

CHAPTER 4

65

4-58. Let ω stand for the angular speed of the disk. Then (32.5 rad/s)(4.0 cm)v Rω= = = 130 cm/s.

The rotation rate in rev/min is (32.5 rad/s)(1 rev/2 radians)(60 s/min)ω π= = 310 rev/min.

†4-59.2

2(9.81 m/s )(200 m) 44.3 m/s.va g v gRR

= = ⇒ = = = Use ω as defined in the solution to

4-58 to find the number of revolutions per minute required to give this

acceleration: 44.3 m/s 60 s 1 rev× × 2.1 rev/min.200 m min 2 radians

vR

ωπ

= = =

4-60. v =3.0 m/s ⇒ 2 ,v vR D

ω = = where D = diameter of hole. For D = 3 mm,

43

2(3.0 m/s) 60 s 1 rev× × 1.9 × 10 rev/min.3 × 10 m min 2 rad

ωπ−= = For D = 25 mm,

33

2(3.0 m/s) 60 s 1 rev× × 2.3 × 10 rev/min.25 × 10 m min 2 rad

ωπ−= =

†4-61. Let T stand for the time for one complete orbit. 2 2

2

2 4 .R v Rv aT R Tπ π= ⇒ = = For an orbit with a

radius of 6500 km and T = 87 min, this gives 2 3

22 2

4 (6500 10 m) 9.4 m/s .(87 min) (60 s / min)

a π ×= =

4-62. (45 rad/s)(0.80 m) 36 m/s.v Rω= = =

†4-63.2

.var

= Let f stand for the number of rotations per second. Then v = 2πrf = 2π × 0.1 m × 1000/s

= 200π m/s. 2 2

6 2(200 m/s) 3.95 × 10 m/s .(0.10 m)

var

π= = = This is = (3.95 × 106 m/s2)/(9.81 m/s2) g

= 4.0 × 105 g.4-64. 1 rev. of blade = π × 0.2 = 0.63 m.

Turning (7000/60) rev/sec, the speed is vtip = 0.63 × 7000 73.3 m/s60

=

The centripetal acceleration is ac = 2tipv /r = (73.3)2/0.1 4 2

= 5.4 × 10 m/s

4-65. v = 0.9999995 c = 0.9999995 × 2.998 × 108 m/s = 2.998 × 108 m/sr = 1.0 km = 1000 ma = v2/r = (2.998 × 108 m/s)2/1000 m 13 2 12

= 8.99 × 10 m/s = 9.16 × 10 gee

4-66. (33.33 rev/min)(1min 60 s)(2 rad rev)(15 cm) 52 cm/s.v R / /ω π= = =4-67. r = 1.50 × 1011 m

v = 2 rTπ =

11 2 × 1.50 × 10 m

365.25 × 24 × 3600π = 2.99 × 104 m/s

ac = 4 2

11

(2.99 × 10 )1.50 × 10

m/s2 3 2 = 5.9 × 10 m/s−

4-68. v = 95 km/h = 26.4 m/s. This is also the tangential speed of a point on the rim of the wheel, so2 2

3 2(26.4 m/s) 2.2 × 10 m/s .(0.32 m)

var

= = =

CHAPTER 4

66

4-69.2 2

2

2 4 .R v Rv aT R Tπ π= ⇒ = = T = 23h 56 min = 86160 s. At the equator,

2 2 62 2

2 2

4 4 (6.37 × 10 m) 3.39 × 10 m/s .(86160 s)eq

RaTπ π −= = = At a latitude of 45°, the radius is the

equatorial radius multiplied by cos 45°, so 2 245 cos 45 2.40 × 10 m/s .eqa a −= ° =

4-70. At the top, 2

.toptop

va

r= 350 km/h = 97.2 m/s.

2(97.2 m/s)500 mtopa = = 18.9 m/s2. At the bottom,

2

.bottombottom

var

= 620 km/h = 172 m/s. 2(172 m/s)

500 mbottoma = = 59.3 m/s2.

At the top, consider the free body diagram for the pilot, who is upside down so thenormal force exerted by the seat points down. Taking the + direction to point down,

Newton’s Second Law says .top topN mg ma+ = topNm

= acceleration the pilot feels =

atop − g = (18.9 − 9.81) m/s2 = 9.09 m/s2. At the bottom, the pilot is upright so thenormal force points up. Taking the + direction to point up, Newton’s Second Law gives

.bottom topN mg ma− = Now bottomNm

= g + abottom = (9.81 + 59.3) m/s2 = 69.1 m/s2.

†4-71. In the table shown, the acceleration was calculated using 2

2

4 .RaTπ=

Planet R (m) T (yr) a (m/s2) 1/R2 (m−2)Mercury 5.79E+10 0.241 0.0395 2.98E-22Venus 1.08E+11 0.615 0.0113 8.57E-23Earth 1.50E+11 1.000 0.00595 4.44E-23

One way to see if the centripetal acceleration is inversely proportional to R2 is to plot a graph of avs 1/R2. The resulting plot is a straight line, proving the proportionality. (In the graph, the valuesof 1/R2 have been multiplied by 1022 to eliminate large negative exponents for the x axis values.)

a vs 1/R2

00.0050.01

0.0150.02

0.0250.03

0.0350.04

0.045

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

1/R2 x 1022 (m-2)

a (m

/s2 )

Another approach to the analysis is to calculate the log of a and R and plot a graph of log a vs logR. If a is inversely proportional to 1/R2, then this graph should be a straight line with a slope of−2.

mgNtop

CHAPTER 4

67

log a−1.4034

−1.94692−2.22548

log R10.7626811.0334211.17609

Log a vs Log R

log a = -1.991(logR) + 20.024

-2.5

-2

-1.5

-1

-0.5

0

10.7 10.8 10.9 11 11.1 11.2

Log R

Log

a

The graph was plotted in a spreadsheet, which was used to find the equation for the resulting line.The slope turns out to be almost exactly −2, again proving the desired proportionality.

4-72. 30 km/h = 8.33 m/sv = v′ + V= −10 j + 8.33 iv = 2 210 8.33 13.0 m/s+ =

The angle θ is θ = tan−1 8.33 4010

⎛ ⎞ = °⎜ ⎟⎝ ⎠

4-73. .ground walk passenger= +v v v If the velocity of the walkway and the velocity of the passenger are inthe same direction, then 1.5 m/s 4.0 m/s 5.5 m/s.ground walk passengerv v v= + = + = If the passenger

is walking in the opposite direction from the walkway’s velocity, the speed of the passenger is1.5 m/s 4.0 m/s 2.5 m/s.ground walk passengerv v v= − = − =

4-74. vrain = −10 j m/s; vcar = 25 i m/sv′ = vrain − vcar = −25 i − 10 j m/sv′ = 2 225 10 26.9 m/s+ =θ = tan−1 25/10 68= °

4-75. vship = 13 i m/svmuzzle = v cos θ i + v sin θ j = 660 cos 20° i + 660 sin 20° j= 620 i + 226 j m/svshot = vship + vmuzzle = (620 + 13) i + 226 j m/s 633 226 m/s= +i j

(magnitude: vshot = 672 m/s, θ = tan−1 226/633 = 19.6°)4-76. v = v′ + V

By the Pythagorean theorem:v = 2 2330 30 329 m/s− =

CHAPTER 4

68

4-77. 2 2 2 2(1.5 m/s) (12 m/s)B W B WV v v= + ⇒ = + = +V v v

V = 12 m/s. 1 1 12tan tan1.5

W

B

vv

θ − −= = = 83 .°

4-78. The speeds add going downstream, so vdown = 6.0 km/h. In 40 min, you can go6.0 km/h × 40 min × 1 h / 60 mind = = 4.0 km. Upstream, the speed of the current must be

subtracted from the speed of the boat, so vup = 1.0 km/h. The time to make the return trip of 4.0

km is 4 km 4.0 h.1 km/hup

dtv

= = =

†4-79. ,train fleaV = +v v where vflea is the velocity of the flea relative to the train. The cat is moving

backwards relative to the train and the flea is moving backwards relative to the cat. The velocityof the flea relative to the train is 0.50 m/s − 0.10 m/s = 0.40 m/s backwards relative to the train.So the flea’s speed relative to the ground is 5.00 m/s 0.40 m/s 4.60 m/s.V = − =

4-80. The position of any point on the shore is given by,d y= +r i j where y is the displacement (positive or

negative) along the y axis of the landing point on theopposite shore. The velocity of the boat relative to theshore is cos ( sin ) .B R Bv v vθ θ= + −V i j The time toreach any point on the opposite shore is given by

,t=r V or ( cos ) ( sin ) .B R Bd y v t v v tθ θ+ = + −i j i j Thisis equivalent to two equations:

( cos )( sin )

B

R B

d v ty v v t

θθ

== −

To reach a point directly opposite the launch point, y = 0, which means sin .R

B

vv

θ = This

describes a right triangle with a hypotenuse vB and an opposite side vR, so we know the adjacent

side is 2 2 .B Rv v− That means 2 2

cos .B R

B

v vv

θ−

= Use this in the first equation to get t:

2 2.

cosB B R

d dtv v vθ

= =−

To find the time to reach any point on the opposite shore, begin with cos ,B

dv t

θ = from which

we deduce 2 2( )

sin .B

B

v t dv t

θ−

= Substitute this into the y equation to get

2 22 2( )

( ) .BR B R B

B

v t dy v v t v t v t d

v t

⎛ ⎞−⎜ ⎟= − = − −⎜ ⎟⎝ ⎠

Rearrange to get 2 2( ) .B Rv t d v t y− = −

Square both sides and rearrange: 2 2 2 2 2( ) 2 ( ) 0.B R Rv v t v yt d y− + − + = The roots of this equation

are 2 2 2 2 2

2 2

2 (2 ) 4( )( ).

2( )R R B R

B R

v y v y v v d yt

v v− ± + − +

=−

The time must be positive, so choose the +

vB

vW

Vθ

y

x

vR

vB

θ

d

CHAPTER 4

69

sign in front of the square root. Rearrange and simplify to get 2 2 2 2 2

2 2

( ).

( )B R B R

B R

v v d v y v yt

v v− + −

=−

As

a quick check, set y = 0 and we get the same result we had in the first part of the problem.

4-81. 30 cm 20 cm (20 cm/s) (13.3 cm/s) .1.5 s 1.5 sesc = + = +v i j i j The rider is walking up with a velocity

30 cm 20 cm (30 cm/s) (20 cm/s) .1.0 s 1.0 srider = + = +v i j i j The rider’s total velocity relative to

the floor is30 cm 20 cm (20 cm/s 30 cm/s) (13.3 cm/s 20 cm/s)1.0 s 1.0 sesc rider= + = + = + + + =V v v i j i j

(50 cm/s) (33.3 cm/s) .+i j The speed is 2 2(50 cm/s) (33.3 cm/s) 60 cm/s.floorv = + = The

direction of the velocity relative to the floor is 1 33.3tan 34 .50

θ −= = °

4-82. v0 = initial velocity relative to the car. If the projectile isobserved to move straight up and down relative to an observeron the ground, then the horizontal component of theprojectile’s initial velocity must have the same magnitude asthe velocity of the car and point in the opposite direction. Thus

0 cos carv vθ = ⇒ 1 1

0

30cos cos50

carvv

θ − −= = = 53.1 .° The maximum height reached is

2 2 20 0

2

( sin ) [(50 m/s) sin 53.1 ]2 2 2(9.81 m/s )

ymax

v vyg g

θ °= = = = 81 m.

4-83. v = v′ + Vv = 2 2

20 + 15 2(20) (15) cos100° 27 km/h− =15/sin θ = 27/sin 100°

sin θ = 1527

sin 100° = 0.547

33θ = °4-84. v = v′ + V,

where v′ = velocity of wind relative to boatV = velocity of boatv = velocity of wind relative to groundv = 2 2

(14) + (32) 2(14) (32) cos 40°− = 23.1 km/hsin

32θ

= sin 40°23.1

θ = sin−1 32 sin 40°23.1

⎛ ⎞⎜ ⎟⎝ ⎠

= 117°

Wind is coming from 13° W of N

θ

v0

vcar

CHAPTER 4

70

4-85. Time of descent t = 15000.5

= 3000 s

(a) Horizontal distance travelled before reaching the ground,

x = 3

360 × 10 (3000) 50 × 10 m60 × 60

⎛ ⎞=⎜ ⎟

⎝ ⎠(b) Relative velocity with respect to still air 60 − 20 = 40 km/h

Therefore x = 3

340 × 10 × 3000 33 × 10 m60 × 60

=

In the upwind direction, the relative velocity is 60 + 20 = 80 km/h, and

x = 3

380 × 10 × (3000) 67 × 10 m60 × 60

=

4-86. sin θ/50 = sin 135°/250

θ = sin−1 1 sin 135°5

⎛ ⎞⎜ ⎟⎝ ⎠

θ = sin−1 (0.1414) = 8°α = 180° − (135° + 8°) = 37°Pilot must point the plane(45° − 37°) = 8° W of Nv/sin α = 250/sin 135°v = sin 37° × 250/sin 135° km/h

213 km/h=

†4-87. v = 2 2 4.2 + 16 2(4.2) (16) cos 70°−

= 15.1 km/hx = 4.2 sin 20° = 1.44 km/hy = (16 − 1.44) km/h = 14.56 km/hsin θ = 14.56/v = 14.56/15.1θ = sin−1 (0.96) 75° (15° E of N)=

4-88. (a) average speed = ∆rrel/∆t = 1 km/2 min = 1 km/(1/30h) = 30 km/h

(b) v = v′ + V with V = 90 km/h; v′ = 30 km/hv = (90 + 30) km/h 120 km/h=

4-89. Let the origin be fixedon car 1. Then in that frame, car 2moves at a speed of 2v0.

CHAPTER 4

71

If they are “together” (at t = 0),then after time t, they areseparated byr = 2 2

0(2 )h v t+

Then,

vrel = drdt

= ( )2 2 204d h v t

dt+ 1/2

= ( )2 2 20

1 42

h v t+ −1/2 ( )208v t

vrel = ( )

20

1/ 22 2 20

4

4

v t

h v t+

= 01/ 22

2 20

2

14

v

hv t

⎛ ⎞+⎜ ⎟

⎝ ⎠[Note: as t → ∞, vrel → 2v0]time (s) rel (m/s)v

10 9.620 17.130 22.340 25.650 27.760 29.170 30.080 30.8

†4-90. vg = v + V(a) 2

gv = 2 2v v−

Total distance acrossand back is 2d.t = dist/vg 2 22 /d v V= −(b) vup = v − Vvdown = v + Vtup = d/(v −V); tdown = d/(v + V)ttot = d/(v −V) + d/(v + V) 2 22 /( )dv v V= −

The trip up and back takes longer (the denominator is larger)

CHAPTER 4

72

†4-91. The velocity of the AWACS relative to theground isvAW = 150 i + 750 j km/hRelative to this, the UFO has velocityvUFO = −950 cos 45° i = 950 sin 45° j km/h =672 i − 672 j km/hThe velocity of the UFO relativeto the ground is

( ) ( )(UFO) 150 750 672 672 km/h= 522 78 km/h

g = + −v i + j i – ji + j

2 2 (UFO) 522 + 78 528 km/hgν = =

The bearing is θ = tan−1 78 8.5° N of W522

⎛ ⎞ =⎜ ⎟⎝ ⎠

4-92. The average vertical velocity is ,yht

= −v j where h is the vertical distance traveled in time t, and

the positive direction is assumed to be up. 2340 m ( 669 m/h) .3.5 hy = − = −v j j No information is

given about the horizontal distance traveled, so there is no way to determine the averagehorizontal component of velocity. Since the speed is 2 2 ,x yv v+ it’s not possible to find the

average speed from the information given.4-93. Assume the airplane is gliding at a constant velocity, so its horizontal

and vertical components of velocity are constant.(a) Taking directions as shown in the diagram,

(240 km/h) cos15 232 km/h.xv = ° = The vertical component ofvelocity is (240 km/h) sin15 62.1 km/h.yv = ° =

(b) The time to fall a distance h = 2000 m = 2 km is given by2 km 0.0322 h 1.9 min.

62.1 km/hy

htv

= = = =

4-94. 2( 25 m/s) (25 m/s) 0 (4.2 m/s ) .12 s

f i

t− − −= = = −

∆v v j ja i j

240 km/h

15°

y

x

vi

vf

CHAPTER 4

73

4-95. (4.2 km/h) cos 20 3.9 km/h.Ev = ° =

(4.2 km/h) sin 20 1.4 km/h.Nv = − ° = −

4-96. (15 km/h) (15 km/h)= +v i j ⇒ 2 2(15 km/h) (15 km/h) (15 km/h) 2 21 km/h.v = + = = The

direction is 1tan 45 ,y

x

vt

vθ −= = ° or 45° E of N.

†4-97. (a) 2 2(6.0 2.0 ) (3.0 2.0 3.0 ) .t t t= + + − +r i j 4.0 ( 2.0 6.0 ) .d t tdt

= = + − +rv i j At t = 2.0 s,

8.0 10 m/s,= +v i j so the speed is 2 2 2 2(8.0 m/s) (10 m/s) 13 m/s.x yv v v= + = + =

(b) 24.0 6.0 m/sddt

= = +va i j at all times. The magnitude of the acceleration is

2 2 2 2 2 2 2(4.0 m/s ) (6.0 m/s ) 7.2 m/s .x ya a a= + = + = The direction relative to the x axis is

1 1 6.0tan tan 56.3 .4.0

y

x

vv

θ − −= = = °

4-98. (a) 2

2 2(1.5 m) 0.553 s.9.81 m/s

htg

= = = Final result: t = 0.55 s.

(b) 0 (60 m/s)(0.553 s) 33 m.x v t= = =

4-99. (a)2 20 sin .max

vyg

θ=2

The maximum possible height would occur for θ = 90°, so

2

2

(700 m/s)(9.81 m/s )maxy = =

2 2.5 × 104m, or 25 km.

(b)202 sin cos .max

vxgθ θ= The maximum horizontal range would occur for θ = 45°, so

2

2

(700 m/s)9.81 m/smaxx = = 5.0 × 104 m, or 50 km.

(c) It is not reasonable to ignore air resistance because at such speeds, air friction would beconsiderable. This is why the rocks do not rise to 25 km or cannot be thrown to distancesapproaching 50 km.

4-100.20

02 sin cos .

2sin cosmax

maxv gxx v

gθ θ

θ θ= ⇒ = For θ = 45°, sin cos 1/ 2,θ θ= = so

20 (9.81 m/s )(180 m) 42 m/s.v = =

4.2 km/h

20°

N

E

CHAPTER 4

74

†4-101.2

0 0 0, .2y x

gty y v t x v t= + − = When y = 0, x = xmax. For θ = 45°, 0 0 0/ 2.x yv v v= = Then the

time to reach y = 0 is

20 0

200 0 0

2422 .

2

v v y gv v y g

tg g

⎛ ⎞ + +⎜ ⎟ + +⎝ ⎠= = (There is another value of t

that solves the equation for y = 0, but it will be negative because the square root in the quadraticformula will be larger than v0. Choose the positive value.) The horizontal range is

2 2 20 0 0 0 0 0 00 4 4

.22 2max

v v y g v v v y gvxgg

⎛ ⎞+ + + +⎜ ⎟= =⎜ ⎟⎝ ⎠

Rearrange: 2 20 0 0 02 4 .maxgx v v v y g− = +

Square both sides: 2 2 2 4 4 20 0 0 0 04 4 4 .max maxg x gx v v v gy v− + = + Cancel the fourth degree terms and

the remaining common factor of 4 and solve for v0:2 2 2

00

(70.87 m) (9.81 m/s )70.87 m 2.0 m

max

max

x gvx y

= =+ +

=

26 m/s.

4-102. For a horizontal initial velocity, the time for the water to fall a distance h is 2htg

= and the

horizontal range is 0 02 .max

hx v t vg

= = This gives 2

09.81 m/s(4.0 m)

2 2(1.0 m)maxgv xh

= = = 8.86

m/s. Assuming that the speed of the water stays the same when the hose is pointed straight up

gives 2 20

2

(8.86 m/s) 4.0 m.2 2(9.81 m/s )maxvyg

= = =

4-103. 90 km/h = 25 m/s. 2 2

2(25 m/s) 8.9 m/s .70 m

var

= = =

4-104. Relative to the ground, the concrete has velocityv = 2ax = 2 × 9.8 × 5 m/s = 9.90 m/sThe velocity of the car is 90 km/h = 25 m/s. If v′ is the velocity of the concrete in the frame of thecar, thenv′ = v − V = −9.9 j − 25 i m/s

2 29.9 25 27m/sv′ = + =

Angle of impact = θ = tan−1 25 689.9

⎛ ⎞ = °⎜ ⎟⎝ ⎠

†4-105. (a) 90 km/h 60 km/h 30 km/h 8.33 m/s.car truckV v v= − = − = =(b) In the reference frame of the truck, the car has to travel a total distance of 90 m: It must travel40 m to catch up to the truck, 10 m to pass the truck, and then travel another 40 m to get ahead of

the truck. The time for the car to travel 90 m at 8.33 m/s is 90 m 10.8 s.8.33 m/s

t = =