Question

A 70.0 kg skier is at rest at the top of a 120m hill. Assuming friction is negligible.
A) What is the speed of the skiers at the bottom of the hill?

Answer 1

Answer:

49 m/s

Explanation:

Initial potential energy = final kinetic energy

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 10 m/s² × 120 m)

v = 49 m/s