A 0.140-kg baseball is dropped from rest. It has a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. The ball is in contact with the ground for 0.0140 s. What is the average force exerted by the ground on the ball during that time?
A) 2.00 N upwards
B) 2.00 N downwards
C) 22.0 N upwards
D) 22.0 N downwards
E) 0 N

Question

Quizplus · Accepted Answer

Using the conservation of energy: 
initial potential energy = final kinetic energy + final potential energy + work done by the ground 
mgh = (1/2)mvf^2 + mgh + Favg*d 
where h = 0 (dropped from rest), vf = 1.00 m/s, d = 2h = 0.280 m 
Solving for Favg, we get Favg = 22.0 N upwards.

A 0140-Kg Baseball Is Dropped from Rest