Question

A ball is thrown horizontally at 11.5 m/s from the top of a hill 45.2 m high. How far from the

base of the cliff does the ball hit the ground?

Answer 1

Vx = 11.5 m/s

h = 45.2 m

g = 9.8 m/s²

Vi.y = 0 m/s (Initial vertical velocity)

x - ?

Solution:

1. h = Vi.y • t + 1/2 • g • t²

Since Vi.y = 0, we will get: h = 1/2 • g • t²

t = sqrt(2h/g)

t = sqrt(2 • 45.2/9.8)

t = 9.22 sec

2. Vx = x/t

x = Vx • t

x = 11.5 • 9.22 = 106.03 meters