Question

A ball, initially rolling at 2.0m/s, rolls down a frictionless hill in 6.0s. At the bottom of the

hill, the ball has a final velocity of 7.0 m/s. It then rolls on a flat, frictionless surface and falls off of an 80m cliff into a lake. Determine
a. The height of the hill.
b. The speed of impact with the lake.

Answer 1

Step-by-step explanation:

a. Energy is conserved. Initially, at the top of the hill, the ball has potential energy and kinetic energy. At the bottom of the hill, the ball has only kinetic energy.

PE + KE = KE

mgh + ½ mv₀² = ½ mv²

2gh + v₀² = v²

h = (v² − v₀²) / 2g

h = [(7.0 m/s)² − (2.0 m/s)²] / (2 × 9.8 m/s²)

h = 2.3 m

b. Using conservation of energy again:

PE + KE = KE

mgh + ½ mv₀² = ½ mv²

2gh + v₀² = v²

v² = (7.0 m/s)² + 2 (9.8 m/s²) (80 m)

v = 40.2 m/s