Question

A projectile is launched straight up at 55.5 m/s from a height of 71 m, at the edge of a sheer cliff. The projectile falls just missing the cliff and hitting the ground below. (a) Find the maximum height of the projectile above the point of firing. m (b) Find the time it takes to hit the ground at the base of the cliff. s (c) Find its velocity at impact. m/s

Answer 1

(a) At the maximum height, final velocity vf = 0

Initial velocity, vi = 55.5 m/s.

Acceleration due to gravity, a = -9.8 m/s^2

Use the expression –

vf^2 = vi^2 + 2 * a * d,

=> 0 = 55.5^2 + 2 * -9.8 * d
=> 19.6 * d = 55.5^2

=> d = 55.5^2 / 19.6 = 157.2 m

So, the maximum height of the projectile above the point of firing = 157.2 m

(b) Let’s use the same equation to determine its velocity at impact. d is the projectile’s vertical displacement.

d = final height – initial height = 0 – 71 = -71 meters

vf^2 = 55.5^2 + 2 * -9.8 * -71
vf = ±?4472

This is approximately 66.9 m/s. But since the projectile is falling, its velocity at impact is negative.

Again, use the following equation to determine the total time it is moving.

vf = vi – g * t
-?4472 = 55.5 – 9.8 * t
t = (-?4472 – 55.5) ÷ -9.8 = 12.5 seconds
Therefore, the total time is approximately 12.5 seconds.

(c) Velocity at impact = -66.9 m/s