Projectile Motion - Horizontal launches

The scale on the left is 1cm = 5m

Draw the positions of the dropped ball every second that it is in the air. Neglect air resistance and assume g = 10 m/s2. Estimate the number of seconds that the ball is in the air

Four positions of a ball thrown horizontally without gravity are shown. Draw the positions of the ball with gravity. Describe the path taken by the ball.

How is motion in the vertical direction affected by motion in the horizontal direction?

They are independent of each other

It is parabolic

Projectile Motion

Rules:

1. Vertical Motion is effected by gravity

2. Horizontal motion is constant

3. Motion in the horizontal direction is independent of the motion in the vertical direction.

4. Assume no air resistance

Physical Expressions

Horizontal Vertical

vHi = vHf = vH

dH = vH t

a = 0 m/s2

vvf = vVi + g t

dv = vvi t + 1/2 g t2

a = g = -9.8 m/s2

vHi = vi cos vvi = vi sin

Example Problem

A projectile is launched from a height of 44.1 m with a initial horizontal speed of 20 m/s. a) How long is it the air? b) how far does it travel horizontally before it hits the ground?

vvi = 0 m/s vH = 20 m/s dv = - 44.1 m g = - 9.8 m/s2

a) dv = vvi t + 1/2 g t2

- 44.1 m = (0 m/s) t + 1/2 (-9.8 m/s2) t2

t = (2 (-44.1 m) / -9.8 m/s2) = 3 s

b) dH = vH t

dH = (20 m/s) (3s) = 60m

20m/s

- 44.1 m

Example Continued

c) What velocity does the projectile hit the ground?

vvi = 0 m/s vH = 20 m/s dv = -44.1 m g = - 9.8 m/s2 t = 3s

vVf = vVi + g t

vVf = (0 m/s) + (-9.8 m/s2) (3s) = - 29.4 m/s

v = (vvf2 + vH

2) = (- 29.4 m/s)2 + (20m/s)2 = 35.6 m/s

= tan-1 (vvf / vH) below the horizontal

= tan-1 (29.4 m/s / 20 m/s)

v = 36 m/s @ 560 below the horizontal

= 55.8 0 below the horizontal

vH

vvfv

Example Continued

vH

vvfvBottom

or… can solve for speed using conservation of energy

KETop + GPETop = KEBottom + GPEBottom

1/2 m vTop2 + (-mghTop) = 1/2 m vBottom

2 + (-mghBottom)

vTop= 20 m/s hTop = 44.1 m hBottom = 0m g = -9.8 m/s2 vBottom = ?

1/2 vTop2 + (-ghTop) = 1/2 vBottom

2 + (-ghBottom)

1/2 (20 m/s)2 + (-(-9.8 m/s2) (44.1m) = 1/2 vBottom2 + (-(-9.8m/s2)(0m)

632.2 m2/s2 = 1/2 vBottom2

vBottom = 2(632.2 m2/s2) = 35.6 m/s