CHAPTER 3PROJECTILE

MOTION

North

South

EastWestpositive x

positive y

negative x

negative y

VECTORS

VECTOR EXAMPLE 1

X-component10

Y-component0

10 m East

10 m

VECTOR EXAMPLE 2

X-component-10

Y-component0

10 m West

10 m

VECTOR EXAMPLE 3

X-component0

Y-component10

10 m North

10 m

VECTOR EXAMPLE 4

X-component0

Y-component-10

10 m South

10 m

COMPONENTS OF A VECTOR

V

VX

VY

x

y

x

y

yx

y

x

VV

VVV

VVθVV

tan

sincos

22

EXAMPLEA wind with a velocity of 40 m/s blows towards 30 NE. What are the x and y components of the wind’s velocity.

smVVsmVV

y

x

/2030sin40sin/6.3430cos40cos

Vx

Vy

V=40

PROJECTILEA projectile is an object with an initial velocity that is allowed to move under the affects of gravity.

Ex. a javelin throw, a package released by an airplane, a thrown baseball

Types of Projectiles

Trajectory: is the path of the projectile.

Because of Earth’s Gravitational pull and their own inertia, projectiles follow a curved path.They have both horizontal and vertical velocities.

Constant horizontal velocity due to inertiaVix = Vfx ax = 0

Constant horizontal velocity due to inertiaViY = 0aY = -9.8 m/s2

IMPORTANT!!!

The ball’s horizontal and vertical motions are completely independent of each other.

CURVED MOTION UNDER GRAVITY1. Ball is simply dropped

2. Ball is thrown horizontally at 10 m/s3. Ball is thrown horizontally at 20 m/s

1 2 3

In the absence of any frictional forces (like air), all the three balls fall to the ground at the same time.The horizontal motion does not affect the vertical acceleration.

All the three balls are pulled to the ground in the same way because of gravity.All the three balls travel the same vertical distance in the same time.

PROJECTILES LAUNCHED HORIZONTALLY

INITIAL VELOCITY

RANGE (DX)

TRAJECTORY

HEI

GH

T (D

Y)

X-component initial velocity horizontal

distance (range)

zero acceleration

final velocity= initial velocity

Y- component zero initial

velocity negative

vertical distance

-9.8 m/s2 acceleration

Negative final velocity

2aDVV

at21tVD

atVV

2i

2f

2i

if

PROBLEM 1A stone is thrown horizontally at 15 m/s from the top of a cliff 44 m high. A) How long does the stone take to reach the ground? B) How far from the base of the cliff does the stone strike the ground? C) Sketch the trajectory of the stone.

X-componentVi = 15 m/sa = 0 m/s2

Vf =15 m/s

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - 44 m

X-componentVi = 15 m/sa = 0 m/s2

Vf =15 m/s

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - 44 mD = Vit + ½ at2

- 44 = ½ (-9.8)t2

t = 3 s

t = 3sD = Vit + ½ at2

D = 15(3) = 45m

A stone is thrown horizontally at a speed of + 5 m/s from the top of a cliff 78.4 m high. A) How long does the stone take to reach the bottom of the cliff? B) How far from the base of the cliff does the stone strike the ground? C) What are the horizontal and vertical velocities of the stone just before it hits the ground?

PROBLEM 2

X-componentVi = 5 m/sa = 0 m/s2

Vf =5 m/s

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - 78.4 m

X-componentVi = 5 m/sa = 0 m/s2

Vf =5 m/s

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - 78.4 mD = Vit + ½ at2

-78.4=½ (-9.8) t2

t = 4 s

t = 4sD = Vit + ½ at2

D = 5(4) = 20 m

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - 78.4 mt = 4 sVf = Vi + at = 0+(-9.8) 4 =- 39.2 m/s 0

1

1

22

7825239

539239

.

.tan

tan

/..5

vVelocityResultant 2fy

2fx

fx

fy

vvsm

θ

How would a) and b) change if the stone is thrown with twice the velocity.Answer:A) would not changeB) would increase two times

PROBLEM 3

How would a) and b) change if the stone is thrown with the same speed but twice the height.Answer:A) would increaseB) would increase

PROBLEM 4

A steel ball rolls with a constant velocity across a table top 0.950 m high. It rolls and hits off the ground + .350 m horizontally from the edge of the table. How fast was the ball rolling?

PROBLEM 5

X-componentVi = ? m/sa = 0 m/s2

Vf =? m/sD = +.350 m

Y- componentVi = 0 m/sa = -9.8 m/s2

D = - .950 m

X-componentVi = ? m/s = Vf

a = 0 m/s2

D = .350 mt = .44 sD = Vit + ½ at2

0.350 = Vi (.44)Vi = .80 m/s

Y- componentVi = 0 m/sa = -9.8 m/s2

D = -0.950 mD = Vit + ½ at2

-0.950= ½(-9.8)t2

t = .44 s

Suppose you throw a ball upward at an angle.The ball has two types of motion

a) Horizontal Motionb) Vertical Motion

The ball moves horizontally at a constant speed. As it moves up gravity slows the ball down (the y speed decreases). At the maximum height, the ball stops. It changes direction and falls downward. Gravity increases the speed of the ball as it falls.

The path of motion is an arc-shaped curve known as a parabola.Horizontal motion has no effect on the time it takes an object to fall to the ground.

PROJECTILES LAUNCHED AT AN ANGLE

LAUNCHINGVELOCITY

LAUNCHINGVELOCITYX-COMP

LAUNCHINGVELOCITYY-COMP

X-component ViX = V cos horizontal

distance (range) zero

acceleration final velocity=

initial velocity

Y- component ViY = V sin @ maximum

height (y velocity = 0)

-9.8 m/s2 acceleration

At the end. D=0 Final

Velocity=Negative

LAUNCHING VELOCITY = v

The initial velocity of a ball in projectile motion is 4.47 m/s. It is projected at an angle of 66 above the horizontal. Find A) how long did it take to land.? B) how high did the ball fly? C) what was its range?

PROBLEM 6

X-componentViX=4.47 cos

660

= 1.8 m/saX = 0 m/s2

VfX =1.8 m/s

Y- componentViY=4.47 sin

660

=4.1 m/saY =-9.8 m/s2

660

4.47 m/sB

A C

X-componentViX=4.47 cos 66

= 1.8 m/saX = 0 m/s2

VfX =1.8 m/s

Y- componentViY=4.47 sin 66

= 4.1 m/saY = -9.8 m/s2

At BVfY = 0 m/s2aY DY = VfY

2 - ViY

2

DY = .86 m

At B (Y-comp)VfY = ViY + aY t0=4.1+(-9.8)t t = .417 s

X-componentViX = 1.8 m/saX = 0 m/s2

t = .83 sDX = ViXt+ ½aX t2

D = (1.8)(.83)D = 1.5 m

ANSWERSTime = .83 sRange = 1.5 m

Height = .86 m

A long jumper leaves the ground at an angle of 20 to the horizontal and a speed of 11 m/s. How far does he jump? What is the maximum height reached?

PROBLEM 7

X-componentViX = 11 cos 200

= 10.33 m/saX = 0 m/s2

VfX = 10.33 m/s

Y- componentViY = 11 sin 200

= 3.76 m/saY = -9.8 m/s2

200

11 m/s B

A C

At BVfY = 0 m/s2aYDY = VfY

2 - ViY

2

DY = .72 m

At B (Y-comp)VfY = ViY + aYt0= 3.76+(-9.8)t t = .38 s

X-componentViX = 11 cos 20

= 10.33 m/s

aX = 0 m/s2

VfX = 10.33 m/s

Y- componentViY = 11 sin 20

= 3.76 m/s

aY = -9.8 m/s2

X-componentViX = 10.33 m/saX = 0 m/s2

t = 2*.38 =.76 sDX = ViXt + ½aXt2

DX =(10.33) (.76)DX = 7.9 m

ANSWERSTime = .76 sRange = 7.9 mHeight = .72 m

A player kicks a football from the ground level with a velocity of 27 m/s at an angle of 30 above the horizontal. Find A) the hang time (the time the ball is in the air) B) the distance the ball travels before it hits the ground. C) its maximum height.

PROBLEM 8

X-componentVi = 27 cos 30

= 23.4 m/sa = 0 m/s2

Vf = 23.4 m/s

Y- componentVi = 27 sin 30

= 13.5 m/sa = -9.8 m/s2

300

27 m/s B

A C

At BVf = 0 m/s2aD = Vf 2 - Vi 2

D = 9.3 m

Vf = Vi + at0=13.5+(-9.8)

(t)t = 1.38 s

X-componentVi = 27 cos 300

= 23.4 m/sa = 0 m/s2

Vf = 23.4 m/s

Y- componentVi = 27 sin 300

= 13.5 m/sa = -9.8 m/s2

X-componentVi = 23.4 m/sa = 0 m/s2

t = 2*1.38 =2.76 sD = Vit + ½at2

D=(23.4) (2.76)D = 65 m

ANSWERSTime = 2.76 sRange = 65.8 mHeight = 9.3 m

A kicker now kicks the ball with the same speed but at an angle of 600 above the horizontal. Find A) the hang time (the time the ball is in the air) B) the distance the ball travels before it hits the ground. C) its maximum height.

PROBLEM 9

X-componentVi = 27 cos 600

= 13.5 m/sa = 0 m/s2

Vf = 13.5 m/s

Y- componentVi = 27 sin 600

= 23.4 m/sa = -9.8 m/s2

600

27 m/s B

A C

At BVf = 0 m/s2aD = Vf 2 - Vi 2

D = 28 m

Vf = Vi + at0=23.4+(-9.8)

(t) t = 2.39 s

X-componentVi = 27 cos 600

= 13.5 m/sa = 0 m/s2

Vf = 13.5 m/s

Y- componentVi = 27 sin 600

= 23.4 m/sa = -9.8 m/s2

X-componentVi = 13.5 m/sa = 0 m/s2

t = 2(2.39) =4.78 sD = Vit + ½at2

D = (13.5) (4.78)

D = 65 m

ANSWERSTime = 5.76 sRange = 65 mHeight = 28 m