Projectile Motion

Projectile Motion

CCHS Physics

Projectile Properties?

Projectile Motion

• Describe the motion of an object in TWO dimensions

• Keep it simple by considering motion close to the surface of the earth for the time being (g = -9.8 m/s2 = constant in y direction)

• Neglect air resistance to make it simpler• Assume the rotation of the Earth has no effect

Projectile Motion

Launch speed = Return Speed. Speed is minimum at apex of parabolic trajectory.

Horizonal component

Net velocity

vx

vertical component

vy v

Above: Vectors areadded in geometricFashion.

Velocity Components at various points of the Trajectory

Projectile Motion

The ball is in free fall vertically and moves at constant speed horizontally!!!

Projectile Motion

Video

What’s wrong with this picture ?

Answer: It never happens ! Only whenthere is no gravity.

Projectile Motion

A History of Projectile Motion

Aristotle:The canon ball travels in a straight line until it lost its ‘impetus’.

Galileo: a result of Free Fall Motionalong y-axis and Uniform Motion along x-axis.

What’s the similarity between a freely-falling ball and a projectile ?

A dropped ball falls in the same time as a ball shot horizontally.Along the vertical, their motions are identical (uniformly accelerated motion (free-fall).

Along the horizontal, notice the ball fired horizontally covers the Same distance in the same unit time intervals (uniform motion along x)

x

yuniform motion

verticalmotion

Projectilemotion

Projectile Motion EquationsHorizontal (x)

x =vxt

Vertical (y)

y=y0 + v0yt−

12

gt2 or dy =12

gt2 + viyt

vy =v0y −gt or g=vfy

−viy

t

1. Along x, the projectile travels with constant velocity. vx=vxo x = vxot

2. Along y, the projectile travels in free-fall fashion.vy = vyo – gt y = vyot – (1/2) gt2 , g= 9.8 m/s2

Projectile motion = a combination of uniform motion along x and uniformly accelerated motion (free fall) along y.

Projectile Motion = Sum of 2 Independent Motions

Everyday Examples of Projectile Motion

1.Baseball being thrown2.Water fountains3.Fireworks Displays4.Soccer ball being kicked5.Ballistics Testing

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.Type IA ball is kicked off a 125 m cliff with a horizontal velocity of 50 m/s. What is the range of the ball?

V = 50 m/s

125

m

Horizontal Verticald = d = vi = vi = a = a = t = t =

-125 m

0 m/s50 m/s

0 m/s2 -9.8 m/s2

5.1 s5.1 s

255 m

d =12

at2 +vit

−125 =1

2−9.8( ) t 2 + 0

d =12

at2 +vit

d =0+50(5.1)d =255 m

t =5.1s

Type IIA ball is kicked of a with a velocity of 50 m/s at an angle of 37° from the horizontal. What is the range of the ball?

Horizontal Verticald = d = vi = vi = a = a = t = t =

0 m

30 m/s40 m/s

0 m/s2 -9.8 m/s2

6.1 s6.1 s

244 m

d =12

at2 +vit

0 =12

−9.8( )t2 + 30t

d =12

at2 +vit

d =0+40(6.1)d =244 m

t =0 or 6.1s

v = 50 m/s37°

v = 50 m/svy = 50sin37 = 30 m/s

37°

vx = 50cos37 = 40 m/s

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

Type IIIA ball is kicked off a 125 m cliff with a velocity of 50 m/s at an angle of 37° from the horizontal. What is the range of the ball?

Horizontal Verticald = d = vi = vi = a = a = t = t =

-125 m

30 m/s40 m/s

0 m/s2 -9.8 m/s2

9.0 s9.0 s

360 m

d =12

at2 +vit

−125 =1

2−9.8( ) t 2 + 30t

d =12

at2 +vit

d =0+40(9.0)d = 360 m

t =9.0 or -2.8s

v = 50 m/s37°

125

mNote, this type

requires the use of the quadratic

equation

Horizontal Range

Maximum Range

• How to maximize horizontal range:– keep the object off the ground for as long

as possible.– This allows the horizontal motion to be a

maximum since x = vxt

– Make range longer by having a greater initial velocity velocity

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.

vi

R

x =R=vxt

R=vi cos ( )t

Range Equation

d =12

at2 + vit

0 =12

gt2 + viyt ⇒ t=

2viy

g=2vi sin

g

What is total t? To solve, set vertical displacement = 0.

R =vi cos ( )2vi sin

g⎛⎝⎜

⎞⎠⎟=

vi2

g2sin cos( )

Trig Identity: 2sincos = sin(2)

R =vi2

gsin 2( ) REMEMBER: ONLY VALID WHEN VERTICAL

DISPLACEMENT IS ZERO (Type II problems).

Projectile Motion• We can see that complementary angles have

the same range because sin = sin2

At what angle do I launch for Maximum Range ?

Need to stay in air for the longest time, and with the fastest horizontal velocity componentAnswer: 45°

Projectile Motion

• What happens when we add air resistance?• Adds a new force on the ball• The force is in the opposite direction to the

ball’s velocity vector and is proportional to the velocity at relatively low speeds

• Need calculus to sort out the resulting motion• Lowers the angle for maximum range

Projectile Motion