Projectile Motion

Question

You throw a stone horizontally at a speed of 5.0m/s from the top of a cliff that is 78.4m high.

A) How long does it take the stone to each the bottom of the cliff?

B) How far from the base of the cliff does the stone hit the ground?

C) What are the horizontal and vertical components of the stone's velocity just before it hits the ground?

Answer 1

In this type of problem, there are 2 components of motion--horizontal and vertical. To solve the problem, we treat them independently.

The typical assumption is that there is no air resistance. This means that the horizontal velocity does not change.

Regardless of the initial horizontal velocity, the vertical motion is "free-fall", meaning that it is strictly a function of gravity. Use this formula for distance:

d = 1/2 * g * t² , where g is the gravitational acceleration--9.8m/s² , and t is the time in seconds.

So:

78.4 = 4.9 * t²

t² = 78.4 / 4.9 = 16

t = 4

The horizontal distance travelled is simply the horizontal velocity times t.

To get the vertical velocity at impact, use v = g * t