Question

A boulder sits atop a steep cliff and someone pushes it off the edge. If the cliff face is 45 meters

high, what is the speed of the boulder just before it hits the ground? (ignore air friction)

Answer 1

See below

Step-by-step explanation:

All of the PE (mgh) will be converted to KE (1/2 mv^2)

mgh = 1/2 m v^2

gh = 1/2 v^2

2 gh = v^2

v = sqrt (2gh) = sqrt ( 2 * 9.81 * 45) = 29.7 m/s