2 – Dimensional 2 – Dimensional KinematicsKinematics

PROJECTILE MOTIONPROJECTILE MOTION

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PROJECTILE MOTIONPROJECTILE MOTION

The trajectory (path) of a projectile is parabolic.The trajectory (path) of a projectile is parabolic.

The vertical motion determines the time of flight.

PROJECTILE MOTIONPROJECTILE MOTION

V0V0YV0Y

V0XV0X

V1

V1X = V0X

V1Y = 0 m/s

V2V2

V2YV2Y

V2X = V0XV2X = V0X

The horizontal motion is uniform. (aX = 0 m/s2)

Vertical motion is accelerated. (aY= g = 9.80 m/s2)

g

SPECIAL CASE ASPECIAL CASE A

Projectile fired horizontally, θ = 0o

Vo

Δdy

Δdx (range)

Case ACase A

Horizontal motion (uniform)Horizontal motion (uniform)

d tx oxv ( cos )ov t d tx ov cos

NOTE : SINCE θ = 0o, cos θ = 1NOTE : SINCE θ = 0o, cos θ = 1

d tx ovWe need to use the vertical motion to find ΔtWe need to use the vertical motion to find Δt

Case ACase AVertical motion (to find Δt)

d t g ty oyv 12

2

Since θ = 0o and voy = vo sin θ, Since θ = 0o and voy = vo sin θ, sin θ = 0 and voy =0sin θ = 0 and voy =0

d g ty 12

2

td

gy

2(time of flight)(time of flight)

Case ACase A

Plug

td

gy

2

t

d

gy

2 into d tx ov d tx ov

d vd

gx oy

2

HOTLINK 1HOTLINK 1

SPECIAL CASE B

Projectile fired from “ground” level, Δdy = 0 m

Vo

Voy

g

Vox Δdx

Case BCase B

Horizontal motion (uniform)Horizontal motion (uniform)

d tx oxv ( cos )ov t d tx ov cos

We need to use the vertical motion to find ΔtWe need to use the vertical motion to find Δt

Case BCase BVertical motion (to find Δt)

d t g ty oyv 12

2

Since Δd = 0 mSince Δd = 0 m

0 12 t v g toy( )

0 012 v g ty

Case BCase B0 0

12 v g ty 0 012 v g ty

12 g t voy

tv

goy

2

(time of flight)(time of flight)tv

go

2 sin

Case BCase B

Plug Plug intointo d tx ov cos d tx ov costv

go

2 sintv

go

2 sin

d vv

gx oo

( )sincos

2

dv

gxo

2 2 sin cos

HOT LINK 2

Case C Case C The most general case.

Δdy

Vo

Vo

Δdy

Case CCase CVertical motion ( to find the time of flight)Vertical motion ( to find the time of flight)

d v t g ty oy 12

2 d v t g ty oy 12

2

Put into standard quadratic formatPut into standard quadratic format

0 12

2 g t v t doy y ( )

And apply the quadratic formulaAnd apply the quadratic formula

tv v g d

goy oy y

2 12

12

4

2

( )( )

( )

t

v v g d

goy oy y

2 12

12

4

2

( )( )

( )

Case CCase CSimplifying, we get :

tv v g d

goy oy y

2 2

t

v v g d

goy oy y

2 2

And since voy = vosinθAnd since voy = vosinθ

tv v g d

go o y

sin sin 2 2 2

t

v v g d

go o y

sin sin 2 2 2

(This is the equation for the time of flight)(This is the equation for the time of flight)

Case CCase C

Horizontal motion (to find the range)Horizontal motion (to find the range)

d v tx oxSince vox = vo cos θ ,Since vox = vo cos θ ,

d v tx o( cos )Now plug the expression for Δt into this equation.Now plug the expression for Δt into this equation.

Case CCase C

dv

gv v g dx

oo o y

cossin sin( ) 2 2 2

The use of “+” or “-” depends on the path of the projectile. The use of “+” or “-” depends on the path of the projectile. 1) If the projectile passes over the apex, use the “-” sign.1) If the projectile passes over the apex, use the “-” sign.

2) If it does not, use the “+” sign.2) If it does not, use the “+” sign.

{Since g is negative, use of the “-” sign gives a greatertime of flight.}{Since g is negative, use of the “-” sign gives a greatertime of flight.}

Case CCase C

For ease of calculation, this is usually written:For ease of calculation, this is usually written:

dv

gv v g dx

oo o y

cossin sin( ) 2 2 2

Use “+” signfor both

(not over apex)

Use “+” signfor both

(not over apex)

Use “-” sign for both(over the apex)

Use “-” sign for both(over the apex)

HOTLINK 3

Homework ProblemHomework ProblemA basketball player shoots a basket by launching the ball at an angle of 60.0o

from a location 1.0 m below the rim at12 m/s. How far is he from the basket

if he makes the shot? (hint : define up as +)

Practice the problems on theProjectile Motion Practice Sheets

The following pagescontain all of thepractice problems.

CHAPTER 7 PROJECTILE MOTION PROBLEMS

1. A ball falls from rest from a height of 490 m. a. How long does it remain in the air? b. If the ball has a horizontal velocity of 2.00 m/s when it begins its fall, what horizontal displacement will it have?

2. An archer stands 40.0 m from the target. If the arrow is shot horizontally with a velocity of 90.0 m/s, how far above the bull’s-eye must he aim to compensate for gravity pulling his arrow downward?

3. A bridge is 176.4 m above a river. If a lead-weighted fishing line is thrown from the bridge with a horizontal velocity of 22.0 m/s, how far has it moved horizontally when it hits the water?

4. A beach ball, moving with a speed of +1.27 m/s, rolls off a pier and hits the water 0.75 m from the end of the pier. How high above the water is the pier?

5. Carlos has a tendency to drop his bowling ball on his release. Instead of

having the ball on the floor at the completion of his swing, Carlos lets go with the ball 0.35 m above the floor. If he throws it horizontally with a velocity of 6.3 m/s, what distance does it travel before you hear a “thud”?

6. A discus is released at an angle of 45° and a velocity of 24.0 m/s. (assume Δdy = 0 m) a. How long does it stay in the air? b. What horizontal distance does it travel?

7. A shot put is released with a velocity of 12 m/s and stays in the air for 2.0 s. (assume Δdy = 0 m)

a. At what angle with the horizontal was it released? b. What horizontal distance did it travel?

8. A football is kicked at 45° and travels 82 m before hitting the ground. a. What was its initial velocity? b. How long was it in the air? c. How high did it go?

9. A golf ball is hit with a velocity of 24.5 m/s at 35.0° above the horizontal. Find :

a. the range of the ball. b. the maximum height of the ball.

10. A car moving at 120 Km/hr on a flat horizontal road loses control and careens off a cliff 75 m high into the valley below. Neglecting wind resistance, how far from the base of this sheer cliff does the car land?

11. You are a detective investigating an accident similar to the one described in problem #1 except that the cliff is 50 m high and the car impacts with the ground 20 m from the base. The speed limit was

50 Km/hr . How fast was the car going when it left the cliff? Could excess speed have contributed to the accident?

12. A baseball is struck with a bat at a height of 1.0 m giving it a speed of

60 m/s at an angle of trajectory of 30o. How far from home plate would the outfielder have to be in order to catch the ball at a height of 1.0 m? (Note: the distance to the outfield wall in this park is 110 m.)

13. A basketball player attempts to shoot a basket from mid-court (13 m from the basket) by making a jump shot and releasing the ball from a height of 3.0 m. If the basket is 3.0 m high and he gives the ball an angle of trajectory of 50o, at what speed must he throw the ball in order to score?

14. A bullet having a muzzle velocity of 150 m/s is fired horizontally from a height of 1.5 m . How much time passes before it strikes the ground? What is its range?

15. A cliff diver in Acapulco launches himself at 1.5 m/s at an angle of 15o above the horizontal from a cliff 60 m above the surf. In order to survive, he must clear the jagged cliffs with a horizontal displacement of 10 m. Will he live to dive again?

THE

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