S SOLUTION STEPS MOTION IN TWO DIMENSIONS For problems involving vector addition or subtraction: 1....

SSOLUTION STEPS

MOTION IN TWO DIMENSIONS

For problems involving vector addition or subtraction:1. Use a protractor and ruler to accurately represent each vector involved in the problem. Use an appropriate scale in representing the vector.A) Graphical Method:Two Vectors: Use the Parallelogram Method and measure both the magnitude and direction of the resultant.Three or more Vectors: Use the Polygon Method and measure both the magnitude and direction of the resultant.B) Trigonometric Component Method:1) break each vector into x and y components.2) use the sign convention and assign a positive sign or a negative sign to the magnitude.3) determine the sum of the x components and repeat for the sum of the y components.4) use the Pythagorean theorem and simple trigonometry to solve for the magnitude and direction of the resultant.

S

For problems involving projectile motion:1. Complete a data table with the information given. Use the proper sign for the quantity represented by the symbol in the data table depending on whether the object was initially moving upward or downward. Use the same sign convention as in free fall problems.2. Use the trigonometric component method to determine the x and y components of the initial velocity.3. Determine which formula or combination of formulas can be used to solve the problem.

SQUAD EQUATION

• READ GIANCOLI P.A-6

SFUNDAMENTALS

• 5. For projectile motion the velocity at the maximum height is zero and the acceleration is g.

• 6. For projectile motion the velocity on the x-axis is constant.

S

Comments: The ball carrier is attached to a Dynamics Cart. The ball is shot straight up when the cart passes a photogate. The ball then follows a trajectory and lands back in the cart. The cart must be moving at a constant velocity for this demonstration to work. This demonstration is more showy if the Dynamics Cart goes through a tunnel after the ball is shot upwards.Name: Ball on a StringCategory: MechanicsSubcategory: 2-D Motion

S

Name: Range of a ProjectileCategory: MechanicsSubcategory: 2-D MotionClasses:

Comments: Use the PASCO projectile launcher to fire a small plastic ball. If the starting and ending heights are the same, the range of the projectile is 2v*sin(2)/g.

SName: The Monkey and the HunterCategory: MechanicsSubcategory: 2-D MotionClasses:

Comments: The blow gun is a long metal tube with a tubing connected on one end and a trigger on the other end. Aim the blow gun by looking down the barrel. The trigger is wired to the electromagnet so that when the marble shoots out of the blow gun the connection between the battery and the electromagnet is broken. The electromagent is a reasonable magnet for a while after the current is cut as well, so it is necessary to weigh the monkey.

SProjectilesA+50+0 Dropped and shot balls hit bench simultaneously. A+50+5 Water projector: Adjustable angle water jet in front of grid.A+50+10 Monkey and the hunter.A+50+15 Reaction jet: "L" tube rotates as water flows through it.A+50+20 Rocket is filled with water and compressed air and launched vertically.A+50+25 Carbon dioxide propelled rocket flies across room on wire.A+50+26 Carbon dioxide propelled rotational device.A+50+35 Ballistics car: Ball ejected from rolling car drops back in.Applet:Newton's cannon: View projectile motion with the earth's curvature taken into account. http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/newt/newtmtn.html

SVectorsA+70+0 X,Y,Z-coordinate system with vector arrows. A+70+5 Vector arrows of various sizes and colors fit in wooden bases.

A+70+10 Relative velocity: Three electric cars on tracks make chalk line.A+70+20 Rope with slug(unit of mass) in center is lifted from ends.A+70+25 Film loop: "Vector addition: Velocity of a boat", 3:35 min.Applet:Manipulate the magnitudes and directions of two vectors, and see the effect this has on their vector cross product. http://www.phy.syr.edu/courses/java-suite/crosspro.html

SThe teacher should locate or create a projectile motion laboratory activity of the following manner – Use an inclined plane to launch a ball bearing or similar object from desk level into a soup can placed on the floor. A popular version of this activity is called “Bull’s Eye.” Remind students to have read this activity the night prior to coming to this class session so that groups and tasks will be predetermined before entering the lab. Have students perform this activity as the teacher circulates and provides assistance where necessary. Allow 10 minutes for clean up of materials. The students have discussed the independence of horizontal and vertical components of motion prior to doing this activity. This activity requires them to predict the landing point of a projectile. It will be apparent which students are prepared and which are not. This lab is very popular. Some teachers have made an addition that they call Bet Your Grade. In this add-on, the teacher sets up the inclined plane and students are permitted to make measurements and calculations for the landing of a ball in a can. They are given only one opportunity to land the ball in the can.

http://home.messiah.edu/~barrett/mpg/cake.mpg

S

• KICK BALL DEMO

SCHAPTER TARGETS

Two-Dimensional Kinematics• be able to use velocity vectors to analyze constant

velocity relative motion.• know how to treat motion with constant velocity/

acceleration in two dimensions.• be able to apply the equations for two-dimensional

motion to a projectile.• be able to calculate positions, velocities, and times for

various types of projectile motion.

SVectors

• General discussion.• Vector - A quantity which has magnitude and

direction. Velocity, acceleration, Force, E Field, Mag Field,

• Scalar - A quantity which has magnitude only. (temp, pressure, time)

• This chapter: We mainly deal withDisplacement D and Velocity = v– Discussion is valid for any vector!

S• Adding vectors in same direction:

SEEMS SIMPLE?

SGraphical Method • For 2 vectors NOT along same line,

adding is more complicated:

Example: D1 = 10 km East, D2 = 5 km North. What is the resultant (final) displacement?

• 2 methods of vector addition:– Graphical (2 methods of this also!)– Analytical (TRIGONOMETRY)

SGraphical Method

• 2 vectors NOT along the same line:

D1 = 10 km E, D2 = 5 km N. Resultant = ?

21

DDDR

21

21

)()( DDDR

21

DDDR

DR = 11.2 km

If D1 is perpendicular to D2

SVector V is 2.50 cm long, scale is 1 cm = 12.2 m/s, find magnitude .

12.22.50

1

mscm

cm

= 30.5 m/s

V

Graphical Method

S

V = V1 + V2

1. Draw V1 & V2 to scale.

2. Place tail of V2 at tip of V1

3. Draw arrow from tail of V1 to tip of V2

This arrow is the resultant V (measure length and the angle it makes with the x-axis)

Graphical Method (TIP TO TAIL)

SGraphical Method (TIP TO TAIL)

Tail

Tip

S

• Consider the vectors AA and BB. Find AA + BB.

AA

BB

AA BB

AA BB

CC = AA + BB

We can arrange the vectors as we want, as long as we maintain their length and direction!!


SGraphical Method • Adding (3 or more) vectors

V = V1 + V2 + V3

S

Graphical Method (Parallelogram)

SGraphical Method • Second graphical method of adding vectors

(parallelogram).

V = V1 + V2

1. Draw V1 & V2 to scale from common origin.

2. Construct parallelogram using V1 & V2 as 2 of the 4 sides.

Resultant V = diagonal of parallelogram from common origin (measure length and the angle it makes with the x-axis)

SCorrect Graphical Method

Tip To Tail

Parallelogram

S Graphically determine the resultant of the following three vector displacements: (1) 34 m, 25º north of east; (2) 48 m, 33º east of north; and (3) 22 m, 56º west of south.

The vectors for the problem are drawn approximately to scale. The resultant has a length of 58 m and a direction 48o north of east. If you actually measured, the actual resultant should be 57.4 m at 47.5o north of east.

34

48

22

S

B

A

B

R = A + B

RR

AB

--BR’R’

AR’ = A - B

Subtraction of Vectors

SSubtraction of Vectors • First, define the negative of a vector:

- V vector with the same magnitude (size) as V but with opposite direction.

Math: V + (- V) 0• For 2 vectors, A - B A + (-B)

SSubtraction of Vectors

S

• Deck of Card Activity

0 degrees from x axis





S• To do the graphical method we have to

measure with ruler and protractor. Takes a lot of time

• We don’t always have an graphic designer on staff to draw our vectors for us.

• So instead we resolve our vectors into its components (x,y)

S

• V is resolved into components: Vx & Vy

V Vx + Vy (Vx || x axis, Vy || y axis)

STrig Functions to Decompose Vector

COS PROJECTES ALONG ONLY X

SIN PROJECTES ALONG ONLY Y Vy = sin V

Vx = cos * V

S

• The magnitude (length) of rr is found using the Pythagorean theorem:

22yx VVv V

rr y

x

SDecomposition Example

d = displacement 500 m, 30º N of E

A coast guard cutter has taken a heading of 30 north of east to find a missing diver. Break up into it’s X and Y

Sm/s 80.6xVm/s, 40.7yV determine the magnitude and direction of the resultant vector

22 2 26.80 7.40 10.0 unitsx yV V V

1 o7.40tan 47

6.80

V

xVy

xyV

The direction is given by an angle of

47o below the positive x-axis.

Vx

Vytan

S An airplane is travelling 735 km/hr in a direction 41.5º of north of west

(a) Find the components of the velocity vector in the northerly and westerly directions. (b) How far north and how far west has the plane travelled after 3.00 h?

o

north 735 km h cos 41.5 550 km hv

o

west 735 km h sin 41.5 487 km hv

north north 550 km h 3.00 h 1650 kmd v t

west west 487 km h 3.00 h 1460 kmd v t

S

cosxv v

sinyv v

112.4

tan15.4

ms

ms

Components are IndependentA ball is thrown 23.0 with a velocity of 22.5 m/s.What is its vertical and horizontal velocity components?

= 22.5 m/s cos 23.0 = 20.1 m/s

= 22.5 m/s sin 23.0 = 8.79 m/sif vy is 12.4 m/s and vx is 15.4 m/s, find the angle .

= 38.8

S

2 2 2c a b 2 2c a b 2 2

3.0 5.0r

km kmv

h h

xv

t

xt

v

1 1500.0

10005.0

kmt m

km mh

xv

t x vt 3.0 0.10

kmx h

h

A ship is steaming north at 5.0 km/h directly across a river that is 5.0 km across. The current is swirling to the east (west to east) at 3.0 km/h. What is its resultant velocity? How far down the bank does it drift at landing?

1. Find resultant v:

2. Find time to cross:

3. Find distance in horizontal direction:

x = 0.30 km

T = 0.100 h

Vr = 5.8 km/h

SVector Multiplication – Cross Product

A--------------

B--------------

x C A B----------------------------

Magnitude of C = ABsin

Direction of C found using right hand rule

Order of cross does matter

vectors MUST be tail to tail

SPhysics applications of vector products

1) Scalar multiple of a vector

Newton’s 2nd Law 2 22 (3 / 4 / )F m a kg m s i m s j ----------------------------

2) Work – Dot Product F--------------

s

cosWork F s

3) Torque on a wheel – Cross Product

F--------------

r

x FTorque r

SMultiplication by a Scalar

• A vector V can be multiplied by a scalar C

V' = C V

V' vector with magnitude CV & same direction as V

I.E a force over a period of time is a greater forceIn relation to the time passed.

SMultiplication by a Scalar

222 034 v smv /525

The vector (4,3,0) has a magnitude:

Vector for velocity has the components from the origin of (4,3,0) After three seconds how far have we moved?

msmsxv 15/53

S“x” and “y” components of motion are independent.

• A man on a train tosses a ball straight up – View this from two reference frames:

Reference frame

on the moving train.

Reference frame

on the ground.

Cart

SConceptual Example 3-6

• Demonstration!!

v0x

S• no air resistance

• ax = 0, vx = constant

• projectile’s weight is only force acting

• ay = g = 9.8 m/s2

• vy = 0 at max height

• x, y motions are independent of each other

• resolve initial velocity into vx, vy and use our basic kinematic equations to solve problem

Projectile Motion

SProjectile Motion

• PHYSICS: y part of motion:vy = gt , y = (½)g t2

SAME as free fall motion!!

An object projected horizontally will reach the ground at the same time as an object dropped vertically from the same point!

(x & y motions are independent)

S

Horizontal:

tvxx xo 0

xox vv

Vertical:

gtvv yoy

20 2

1gttvyy yo

ygvv yoy 222

SINCE G IS ALWAYS DOWN, MAKE IT NEGATIVE FOR ALL PROJ MOTION AND POSITIVE FOR FREE FALL

(no acceleration)

gtvv o

2

2

1gttvy o

tvx 0

ygvv o 222

Your book says negative, just remember in most cases g will be opposite the initial velocity (so -9.8 m/s) (Except free fall)

SBall Rolls Across Table & Falls Off

vy=gt

vyo=0

In this case I take down as positive since vyo=0

S

• Summary: Ball rolling across the table & falling.

• Vector velocity v has 2 components:

• Vector displacement D has 2 components:

2

2

1gttvy o tvx x0

For problems like this if we set y at the top = 0 then gravity is – 9/8 m/s2

gtvy 0xx vv

SA rookie soldier in basic training jumps over an obstacle, and fires a shot from his rifle at the same time. The barrel of the rifle is exactly horizontal at the moment of the shot.(no air res.)Who touches the ground first, the soldier or the bullet?

The soldier and the bullet touch down at the same time.

SDriving off a cliff!!

y is positive upward, y0 = 0 at top. Also vy0 = 0

• How fast must the motorcycle leave the cliff to land at

x = 90 m, y = -50 m? vx0 = ?

019.3

90v

(no acceleration)

0vt

x

negative y

0/2.28 xvsm

2

2

1gty t

g

y

2

ts 19.3

tsm

m

2/8.9

)50(2

S

21

2y gt 2 2y

tg

2yt

g

2

2 25.0

9.80

mt

ms

A flag pole ornament falls off the top of a 25.0 m flagpole. How long would it take to hit the ground?

t = 2.26 s

Now assume that a bullet causes it to Fall with a horizontal velocity of 5 m/sHow far does it travel from the pole?

X = v x t

X= 5m/s x 2.26s = 11.3 m

S

21

2y gt

2yt

g

2

2 40.0

9.80

mt

ms

15.0 2.86m

x ss

A stone is thrown horizontally from the top of a cliff that is 40.0 m high. It has a horizontal velocity of 15.0 m/s. We want to find how long it takes the stone to fall to the deck and how far it will travel from the base of the cliff.

Plug in the given values:

t =2.86 s

Find x :x = vx t

x = 42.9 m

S

1000 1375

1 3600x

km m hv

h km s

x

xt

v

15 250

104t m

ms

21

2y gt

2

2

19.80 50.5

2

ms

s

B-17 is flying level at 375 km/h. The bombs travel a horizontal distance of 5250 m. What was the altitude of the plane at the time they called the old "bombs away"?

x = vxt

t = 50.5 sFind vertical distance (altitude) to take 50 sec:

y = 12 509 m

= 104 m/s

S

0 1.8 m sxv 0xa 0 0yv 29.80 m sya

0 0y 3.0 st

22 21 10 0 2 2

0 0 9.80 m s 3.0s 44 my yy y v t a t y

1.8 m s 3 s 5.4 mxx v t

A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and 3.0 s later reaches the water below. 1) How high was the cliff, and 2)how far from its base did the diver hit the water? Choose downward to be the positive y direction. The origin will be at the point where the diver dives from the cliff.

In the horizontal direction, In the vertical direction

and

The distance from the base of the cliff to where the diver hits the water is found from the horizontal motion at constant velocity:

S

0 0yv

0 0y

29.80 m sya

36.0 m 22.2 m s 1.62 sx xx v t t x v

22 21 10 0 2 2

0 0 9.80 m s 1.62 s 12.9 my yy y v t a t y

,

The vertical displacement, which is the height of the building, isfound by applying.

mx 0.36

smvx /2.22

20 2

1gttvyy yo

A ball thrown horizontally at 22.2 m/s from the roof of a building lands 36.0 m from the base of the building. How tall isthe building?

SThe pilot of an airplane traveling 180 km/hr wants to drop supplies to flood victims isolated on a patch of land 160 m below.

The supplies should be dropped how many seconds before the plane is directly overhead?

0 0yv 29.80 m sya 0 0y 160 my

The time of flight

Note that the speed of the airplane does not enter into this calculation.

2

2

1gty t

sm

m

2/8.9

)160(2

ts 71.5

200 2

1gttvyy

SProjectile Motion

S

• PLAY JB 5 CLIP HERE

SProjectile Motion• General Case:

Vxo=Vocos

Vyo=Vosino

a = g (down always)

SIf Bubba is traveling at 20m/s and the ramp is 30 deg will he make a ramp distance of

http://www.youtube.com/watch?v=_GAGjkkFBv4&feature=related&safety_mode=true&persist_safety_mode=1&safe=active

S

θ0 = 37º, v0 = 20 m/s

vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s

a. Max height? b. Time when hits ground?

c. Total distance traveled in the x direction?

d. Velocity at top? e. Acceleration at top?

The football leaves punters foot at angle of 37o and 20.0 m/s.

S

2 20 2y yv v gy

202 ygy v

20

2yv

yg

2

2

20.7

2 9.80y

ms

dms

2

2

2

428.5

19.60y

ms

dms

.

A ball is given an initial velocity of 22. 7 m/s at an angle of 66.0 to the horizontal. Find how high the ball will go?

vy0 = v0 sin 0 vy0 = 22.7 m/s (sin 66.0) vy0 = 20.7 m/s

Find y: vy0 = 20.7 m/s and vy = 0.

y = 21.9 m

2 possible Formulas

S

t

dv x

x vm

sx 48

4 5. st 5.4

2

1

0y yv v gt 00 yv gt 0yv gt

29.80 2.25i

mv s

s

01

0

tan y

x

v

v

1 22.05tan

10.67

m s

m s

The biker has a hang time of 4.5 s. He lands 48 m from the ramp, (assume no ramp height) what was the ramp angle?

vx0 = 10.67 m/s

t = 2.25 s

Vy0 = 22.05 m/s

= 64

NEED TO FIND Vy0 and Vx0

NEED TO FIND t at max height

NEED TO FIND t at max height

NEED TO FIND Vy

S• A confused soldier fires his artillery piece at angle that maximizes

Vy. His enemy utilizes an angle of 40o. If both are in the air for 3 seconds (same height) what was the velocity of the enemy's ball.

gtvV yy 0

At Vy Max the Vxo =0, t to the top = 1.5s

gtVyo

smVy /8.22)40(sin/7.14 yvsm smVy /7.140

smVy /7.140

)5.1)(2/8.9( ssmVyo

Since both are in the air for the same amount of time, both have same Vyo

40

y

S

0 345 sin 32.0y

mv

s

0y yv v gt 0 0cosx v t

345 cos32.0 37.27m

x ss

A naval gun fires a 16 inch projectile. The muzzle velocity (speed of the bullet) is 345 m/s with an angle of 32.0 . What is the range of the shot?

t = 37.27 s

x = 10904 m

Find the vertical velocity:

Find the time:Find range:

smvy /8.1820

Hint since both are at y=o then you don’t need the quadratic to solve for time

uptsm

sm

2/8.9

)/8.182(

totalts 2)6.18(

upts 6.18

S

0y yv v gt 0yv gt 0 2

1.559.80

2y

m sv

s

0

0

tan y

x

v

v 0

0 tany

x

vv

0

7.60

tan 22.2x

msv

0xx v t 18.62 1.55m

x ss

You throw a potato at an angle of 22.2 . If it is in the air for 1.55 s, how far did it go?

m/s 18.62 0 xv

m/s 7.60 0 yv

Find the vertical velocity:

Find the horizontal velocity:

m 28.9 x

Find the horizontal displacement:

S45

104.50x

x m mv

t s s

29.8 4.5

2 22.052 2o o

msat msv v at v v at v at v

s

2 22 2 22.05 10 24.21y x

m m mv v v

s s s

o22.05

tan 65.610

y

x

mv s

mvs

A ball is thrown at some angle. The ball is in the air for 4.50 seconds before it hits. If it travels 45.0 meters before it hits the ground, what was the initial velocity of the ball (magnitude and direction please)?

S

But what if the height is not the same after landing?

SA husky kicker wants to provide as much hang time as possible For his coverage team to get down the field. If he kicks the ball with a velocity 40m/s at an angle of 35 degrees above horizontal, how much time does his team have to get down field and pound a cougar?At what distance from kick will we find a pounded cougar returner?

vy0 = v0 sin vx0 = v0 cos vy0 = 40 sin 35 vx0 = 40 cos35

upyy gtvV 0

mx 4.153

uptsmsm )/8.9()/9.22(0

uptsm

sm

2/8.9

)/9.22(stup 34.2

ssttotal 68.4)2)(34.2(

)(35cos tvx )68.4(7.32 sx

35vy0 = 22.9 m/s vx0 = 32.7 m/s

Hint since both are at y=o then you don’t need the quadratic to solve for time

SA football leaves a punters foot 1.0 m above the

ground at 20.0 m/s. θ0 = 37º. Find t then x.

2

2

1gttvyy oo

Vy0 = 12.0 m/s, θ0 = 37º

mts

mt

s

m0.1129.40 2

2

22)8.9(

2

10.120.10 t

s

mt

s

mmm

Quadratic Equation

sst 53.2,081.0

t

xvx

0 53.2

/16x

sm

xm 5.40

S

Quadratic Equation45 .0 m

x

y

A stone is thrown off the top of a building from a height of 45.0 m. The stone has a launch angle of 62.5 and a speed of 31.5 m/s.

How long is the stone in flight

What is its speed just before it hits the ground?

What angle does it hit? 5.62sin5.310 yV

stup 84.2

))(/8.9(/9.270 2uptsmsm

gtvv yy 0

smVy /9.270

5.62cos5.310 xV

xfx vv 0smvxf /5.14

gyvv f 220

2 )45)(/8.9(2)/9.27( 222 msmsmv f

22 /1666 smv f smv f /8.40

Vf at bottom from 45 meters

sstarch 68.5)2)(84.2(

S

smv /2.4322 5.148.40 v22

xy vvv

st down 32.12/1

))(/8.9(/9.27/7.40 2/12

downtsmsmsm

gtvV yy 0

x

y

v

vtan

sm

sm

/5.14

/7.40tan

)8.2(tan 1

What is its speed just before it hits the ground?

What angle does it hit?

o3.70

yv

xv

Remember that this 2.8 is negative (so will be below horizon)

ssttotal 0.768.532.1

SFORMULAS GIVEN ON AP

SEXAM TIME

S• Ichiro clobbers a fastball toward Safeco center-field. The ball is hit at 1 m (yo ) above the plate, and it’s initial velocity is 36.5 m/s (v) at an angle of 30o above horizontal. The center-field wall is 113 m from the plate and is 3 m (y) high.

• How high does it fly?• What time does the ball reach the fence?• At What height does it clear the fence? • Does he get a home run?

vv h

D

y0

X0

Note: t to wall is not necessarily twice the tup

vv

v0x

v0y

S• How high does it fly ?

30sin)/5.36( smvyo

))(/8.9(/25.180 tsmsm stup 86.1

0y yv v gt

smvyo /25.18

22 )86.1)(/8.9(2

1)86.1)(/25.18(1 ssmssmy mmmy 95.1695.331

2

2

1gttvyy yoo

my 0.18

h

L

y

x

vv00

P

S• How long does it take for ball to reach the wall ?

smvxo /6.31

2

2

1gttvyy yoo

30cos)/5.36( smvxo

t

xvx

0

wallt

msm

113/6.31 stwall 57.3

•What is the balls height at the wall ?

2)57.3)(/8.9(2

1)57.3)(/25.18(1 smssmmy

my 7.3

Not the total time of flight!!!!

mmmy 45.6215.651

S

A depth charge is shot from the top of an aircraft carrier125 m above sea level with an initial speed of 65.0 m/s at an angle of 37.0º with the horizontal(a) Determine the time taken by the projectile to hit point P at

ground level. (b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before projectile hits point P. (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal.(f) Find the maximum height above the cliff top reached

S0 65.0 m sv ya g

0 125y 0 0 0sinyv v smvyo /6.3537sin65

2

2

1gttvyy yoo

22 )/9.4()/6.35(1250 tsmtsmm

sst 4.10,45.2

(a) Time to hit P

4 5 .0 m

x

y

S

2 22 2 51.9 m s 63.1m s 81.7 m sx yv v v

(d) The magnitude of the velocity is found from the x and y components calculated in part c) above.

o 2

0 0 0sin 65.0 m s sin 37.0 9.80 m s 10.4 s

63.1m s

y yv v at v gt

t

xvx

0 )4.10(/3.54 ssmx mx 565

0.37cos)/0.65( smvxo sm /3.54

(b) Determine the range

(c) the horizontal and the vertical components of its velocity, not where it is abeam launch site. Hint: Just find Vy at 10.4s. Vx is same.

o

0 0cos 65.0 m s cos 37.0 51.9 m sxv v .

S1 1 o63.1

tan tan 50.651.9

y

x

v

v

o50.6 below the horizon

2 2 2 2

0 0 0 0 max

2 2 o2 2

0 0max 2

2 0 sin 2

65.0 m s sin 37.0sin78.1 m

2 2 9.80 m s

y y yv v a y y v gy

vy

g

(e) the angle made by the velocity vector with the horizontal.

-63.1

51.9

Can use y=Vot+2gy as well

S1-D Velocity Trick #3“I Hate the Quadratic Approach”

• Velocity at beginning is + Vyo and Velocity at landing is –Vyo

• So we can figure out the time by subtracting the time it takes gravity to move us from Vyo to –Vyo

0y yv v gt tsmvv yy )/8.9(00 + -

tsm

vv yy

200

/8.9

If both points are at a y=0 on path

• Velocity at beginning is + Vyo and Velocity at landing is –Vyo

• So we can figure out the time by subtracting the time it takes gravity to move us from Vyo to –Vyo

If both points are at a y=0 on path

S31. A projectile is shot from the edge of a cliff 125 m above ground

level with an initial speed of 65.0 m/s at an angle of 37.0º with the horizontal

(a) Determine the time taken by the projectile to hit point P at ground level.

(b) Determine the range X of the projectile as measured from the base of the cliff. At the instant just before projectile hits point P.

(c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal.(f) Find the maximum height above the cliff top reached

0 65.0 m sv ya g0 125y 0 0 0sinyv v

smvyo /6.3537sin65 2

2

1gttvyy yoo

22 )/9.4()/6.35(1250 tsmtsmm

sst 4.10,45.2

(a) Time to hit P

4 5 .0 m

x

y

S

2 22 2 51.9 m s 63.1m s 81.7 m sx yv v v

(d) The magnitude of the velocity is found from the x and y components calculated in part c) above.

o 2

0 0 0sin 65.0 m s sin 37.0 9.80 m s 10.4 s

63.1m s

y yv v at v gt

t

xvx

0 )4.10(/3.54 ssmx mx 565

0.37cos)/0.65( smvxo sm /3.54

(b) Determine the range

(c) the horizontal and the vertical components of its velocity, not where it is abeam launch site. Hint: Just find Vy at 10.4s. Vx is same.

o

0 0cos 65.0 m s cos 37.0 51.9 m sxv v .

S1 1 o63.1

tan tan 50.651.9

y

x

v

v

o50.6 below the horizon

2 2 2 2

0 0 0 0 max

2 2 o2 2

0 0max 2

2 0 sin 2

65.0 m s sin 37.0sin78.1 m

2 2 9.80 m s

y y yv v a y y v gy

vy

g

(e) the angle made by the velocity vector with the horizontal.

-63.1

51.9

SLeading the Reciever

• Demonstration!!

SShooting the Monkey(tranquilizer gun)

Where does the zookeeper aim if he wants to hit the monkey?

( He knows the monkey willlet go as soon as he shoots ! )

NOTES

S2 derivations

• vertical displacement as a function of horizontal range – shape of trajectory is inverted parabola

• horizontal range as a function of launch angle

SWHICH ONE IS CORRECT?

SShooting the Monkey...

If there were no gravity, simply aim

at the monkey

r = r0

r =v0t

SShooting the Monkey...

rr = vv0 t - 1/2 gg t2

With gravity, still aim at the monkey! rr = r0 - 1/2 gg t2

Dart hits the monkey!

Both Ball And Monkey Fall at the Same Rate

SRecap:Shooting the monkey...

x x = = xx00

yy = -1/2 gg t2

This may be easier to think about.

It’s exactly the same idea!!

xx = = vv0 0 tt

yy = -1/2 gg t2

S32. A shot-putter throws the shot with an initial speed of15.5 m/s at a 34.0º angle to the horizontal. Calculate the horizontal distance travelled by the shot if it leaves the athlete’s hand at a height of 2.20 m above the ground.

34.015.5 m

/s

2.2m

o

0 15.5sin 34.0 m s 8.67 m syv

29.80 m sya

,

,

my 20.20 my 0sin0vvyo

2

2

1gttvyy yoo 29.4)/67.8(2.20 ttsmm

sst 99.1,225.0

t

xvx

0

)99.1(/3.13 ssmx mx 6.25

sin0vvxo 0.34cos)/5.15( smvxo sm /3.13

SMAX RANGE

• For a given range, R < Rmax there are 2 different firing angles that will land the projectile in the same place. If you fire a shell at 30 degrees, it will land at exatly the same spot as if you fire it at 60 degrees.

• The 2 angles are related to each other by 2 = 90 – 1, so at Rmax, 2= 1 = 45o

g

VR oo 2sin2

SExample 3-8

• Range (R) of projectile Maximum horizontal distance before returning to ground. Derive a formula for R.

S• Range R the x where y = 0!• Use vx = vx0 , x = vx0 t , vy = vy0 - gt y = vy0 t – (½)g t2, (vy) 2 = (vy0)2 - 2gy

• First, find the time t when y = 0 0 = vy0 t - (½)g t2 t = (2vy0)/g

t = 0 (of course!) and t = (2vy0)/g • Put this t in the x formula: x = vx0 (2vy0)/g R

R = 2(vx0vy0)/g, vx0= v0cos(θ0), vy0= v0sin(θ0) R = (v0)2 [2 sin(θ0)cos(θ0)]/g

R = (v0)2 sin(2θ0)/g (by a trig identity)

MAX RANGE

S19.(II) A fire hose held near the ground shoots water at a speed of 6.8m/s. At what angle(s) should the nozzle point in order that the water land 2.0 m away? Why are there two different angles? Sketch the two trajectories.

2

0 0

2

0 22

0

1 o o

0 0

sin 2

2.0 m 9.8 m ssin 2 0.4239

6.8 m s

2 sin 0.4239 13 ,77

vR

g

Rg

v

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2

SA flaming load is catapulted at an angle of 40o from the horizontal. With what speed must the projectile be launched so that it lands 1500m from the launching point?

077. vvxo

40sin0vvyo

064.0 vvyo

40cos0vvxo

ovt 065.ts

mv

s

m)8.9(64.0

20 gtvV yy 0Find t at zenith

Total time in flight is the double of the time in flight for the zenithovt 13.

t

xvx

0

ov

mv

13.

150077. 0 mvv o 1500)13)(.77(. 0

1500)10.0( 2 ov

15000ov s

mvo 4.122

40o

1500

Vy0

Vxo

NEED TO FIND t through Vy0

S

Ch. 3 p.65 : 4, 5, 6, 9,10, 18, 19, 20, 21, 23, 27, 30, 31, 32, 39,40, 41, 44, 45

S30. (II) A projectile is fired with an initial speed of sm 2.65

at an angle of 34.5º above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the velocity of the projectile 1.50 s after firing.

0 0 0sinyv v

2 2

0 02y y yv v a y y

0yv

22 2 2 o2 2 2 20 0 0 0 0

max 2

65.2 m s sin 34.5sin sin0 69.6 m

2 2 2 2 9.80 m s

y y

y

v v v vy

a g g

2 21 10 0 0 02 2

o

0 0

2

0 sin

2 65.2 m s sin 34.52 sin7.54 s and 0

9.80 m s

y yy y v t a t v t gt

vt t

g

o

0 0cos 65.2 m s cos 34.5 7.54 s 405 mxx v t v t

o

0 0cos 65.2 m s cos 34.5 53.7 m sv

o 2

0 0 0sin 65.2 m s sin 34.5 9.80 m s 1.50s 22.2 m sy yv v at v gt

. (a) The maximum height is found from Eq. 2-11c,

, with

at the maximum

height.

(b)The total time in the air is found from Eq. 2-11b, with a total vertical displacement

of 0 for the ball to reach the ground.

The time of 0 represents the launching of the ball.(c)The total horizontal distance covered is found from the horizontal motion at constant

velocity.

(d)The velocity of the projectile 1.50 s after firing is found as the vector sum of the horizontal

and vertical velocities at that time. The horizontal velocity is a constant

. The vertical velocity is found from

Eq. 2-11a.

S

VECTORS

SExample 1: Find the height of a building if it casts a shadow 90 m long

and the indicated angle is 30o.

90 m

300

The height h is opposite 300

and the known adjacent side is 90 m.

h

h = (90 m) tan 30o

h = 57.7 mh = 57.7 m

0tan 3090 m

opp h

adj

SFinding Components of Vectors

A component is the effect of a vector along other directions. The x and y components of the vector (R, are illustrated below.

x

yR

x = R cos

y = R sin

Finding components:

Polar to Rectangular Conversions

SExample 2: A person walks 400 m in a direction of 30o N of E. How far is the displacement east and how far north?

x

yR

x = ?

y = ?400 m

E

N

The y-component (N) is OPP:

The x-component (E) is ADJ:

x = R cos y = R sin

E

N

SExample 2 (Cont.): A 400-m walk in a direction of 30o N of E. How far is the displacement east and how far north?

x = R cos

x = (400 m) cos 30o

= +346 m, E

x = ?

y = ?400 m

E

N Note:Note: x is the side x is the side adjacent to angle adjacent to angle

303000

ADJ = HYP x Cos ADJ = HYP x Cos 303000

The x-component The x-component is:is:RRxx = = +346 m+346 m


y = R sin

y = (400 m) sin 30o

= + 200 m, N

x = ?

y = ?400 m

E

N

OPPOPP = HYP x = HYP x SinSin 303000

The y-component The y-component is:is:RRyy = = +200 m+200 m

Note:Note: yy is the side is the side oppositeopposite to angle to angle

303000


Rx = +346 m

Ry = +200 m

400 m

E

NThe x- and y- The x- and y- components components are are eacheach + in + in

the first the first quadrantquadrant

Solution: The person is displaced 346 m east and 200 m north of the original

position.

S

RELATIVE VELOCITY

NOT ON THE AP EXAM

SSection 3-8: Relative Velocity

SSection 3-8: Relative Velocity• A useful example of vector addition!

• Example: 2 trains approaching each other (along a line) at 95 km/h each, with respect to the Earth.

• Observers on either train see the other coming at 95 + 95 = 190 km/h. Observer on ground sees 95 km/h.

Velocity depends on reference frame!!

SVelocities not along the same line

• Need to use full vector addition.– A common error is adding or subtracting wrong

velocities– A method to help avoid this is:

Proper subscript labeling of velocities

• CONVENTION: – Velocities with 2 subscripts. First = object, O,

Second = reference frame, R.

vOR

SConceptual Example 3-10:

Boat Crossing A River

• vBS = vBW + vWS

• Outer subscripts on both sides are the same!• Inner subscripts are the same!

SCan extend this to more than 2

v’s • Suppose, to the previous example, we add

a fisherman walking on boat with velocity

vFB = velocity of the Fisherman with respect to the Boat:

vFS = vFB + vBW + vWS

• Outer subscripts on both sides are the same!• Inner subscripts are the same!

• Finally: Relative velocities obey:

vAB = -vBA

SExample 3-11

SExample 3-12

Ssm 30.2

s,m 20.1*39. (II) A boat can travel

in still water. (a) If the boat points its prow directly across a stream whose current is

what is the velocity (magnitude and direction) of the boat relative to the shore? (b) What will be the position of the boat, relative to its point of origin, after 3.00 s? (See Fig. 3–30.)

2 2 2 2

boat rel. water rel. boat rel.shore shore water

1.20 2.30 2.59 m sv v v

1 o o o1.20tan 27.6 , 90 62.4 relative to shore

2.30

boat rel.water

v

water rel.shore

v

boat rel.shore

v

39. Call the direction of the flow of the river the x direction, and the direction the boat is headed the y direction.

(a)

S*40. (II) Two planes approach each other head-on. Each has a speed of

h,km 785

and they spot each other when they are initially 11.0 km apart. How much time do the pilots have to take evasive action?

40. If each plane has a speed of 785 km/hr, then their relative speed of approach is 1570 km/hr. If the planes are 11 km apart, then the time for evasive action is found from

11.0 km 3600 sec 25.2 s

1570 km hr 1 hr

dd vt t

v

Sh.km 600

hkm 100

*41. (II) An airplane is heading due south at a speed of

If a wind begins blowing from the southwest at a speed of

(average), calculate: (a) the velocity (magnitude and direction) of the plane relative to the ground, and (b) how far from its intended position will it be after 10 min if the pilot takes no corrective action. [Hint: First draw a diagram.]

o o

2 2

plane rel.ground

1 o

0, 600 km h 100 cos 45.0 ,100sin 45.0 km h

70.7, 529 km h

70.7 km h 529 km h 540 km h

70.7tan 7.6 east of south

529

v

16

100 km h h 17 kmxx v t

planerel. air

v

air rel.ground

v

plane rel.ground

v

41. Call east the positive x direction and north the positive y direction. Then the following vector velocity relationship exists.(a)

(b) The plane is away from its intended position by the distance the air has caused it to move. The wind speed is 100 km/h, so after 10 min (1/6 h), the plane is off course by

Ssm 50.1

sm 50.0

44. (II) A passenger on a boat moving at

on a still lake walks up a flight of stairs at a speed of

(Fig. 3–39). The stairs are angled at 45º pointing in the direction of motion as shown. What is the velocity of the passenger relative to the water?

passenger passenger boat rel.rel. water rel. boat water

o o 0.50cos 45 ,0.50sin 45 m s

1.50,0 m s 1.854,0.354 m s

v v v

2 2

passengerrel. water

1.854 0.354 1.89 m sv

boat rel.water

v

passengerrel. boat

v

passengerrel. water

v

44. Call the direction of the boat relative to the water the x direction, and upward the y direction. Also see the diagram.

S*45. (II) A motorboat whose speed in still water is

sm 60.2

must aim upstream at an angle of 28.5º (with respect to a line perpendicular to the shore) in order to travel directly across the stream. (a) What is the speed of the current? (b) What is the resultant speed of the boat with respect to the shore? (See Fig. 3–28.)

o28.5

boat rel.water

2.60 m sv

o

water rel. boat rel. water rel.shore water shore

sin 2.60 m s sin 28.5 1.24 m sv v v

o

boat rel. boat rel. boat rel.shore water shore

cos 2.60 m s cos 28.5 2.28m sv v v

boat rel.water

v

water rel.shore

v

boat rel.shore

v

water rel.shore

v

boat rel.shore

v

boat rel.water

v

110 m

260 m

o45

45. Call the direction of the flow of the river the x direction, and the direction straight across the river the y direction. The boat is traveling straight across the river. The boat is headed at

upstream, at a speed of

.(a)

(b)

SSuppose a cannon is placed at the bottom of a hill that has a slope of 15o with the horizontal.The cannon itself is at 45o with the vertical. The cannon is fired and the cannonball as anInitial v of 125 m/s. Where will cannonball strike the hill?

S

• Two footballs are thrown from the same point on a flat field. Both are thrown at an angle of 30o above the horizontal. Ball 2 has twice the initial speed of ball 1. If ball 1 is caught a distance D1 from the thrower, how far away from the thrower D2 will the receiver of ball 2 be when he catches it?

(a) D2 = 2D1 (b) D2 = 4D1 (c) D2 = 8D1

CONCEPTUAL QUESTION

S The distance a ball will go is simply x = (horizontal speed) x (time in air) = v0x t

To figure out “time in air”, consider the equation for the height of the ball:

When the ball is caught, y = y0

0tg2

1tv 2

y0

0tg2

1vt y0

t 0 (time of throw)

(time of catch)

two solutions

2y00 tg

2

1tvyy

g

vt y02

SLecture 2, Act 3Solution

g

v2t y0 So the time spent in the air is proportional to v0y :

Since the angles are the same, both v0y and v0x for ball 2are twice those of ball 1.

ball 1ball 2

v0y ,1

v0x ,1

v0y ,2

v0x ,2

v0,1

v0,2

Ball 2 is in the air twice as long as ball 1, but it also has twice the horizontal speed, so it will go 44 times as far!!

x = v0x t

S SOLUTION STEPS MOTION IN TWO DIMENSIONS For problems involving vector addition or subtraction: 1....

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Transcript of S SOLUTION STEPS MOTION IN TWO DIMENSIONS For problems involving vector addition or subtraction: 1....