Question

A ball of mass 0.15 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.960 m. b). (Using conservation of energy on the way up) What is the velocity of the ball right after the collision?

Answer 1

Answer:

Explanation:

If v be the velocity just after the rebound

Kinetic energy will be converted into potential energy

1/2 m v² = mgh

v² = 2gh

v = √ 2gh

= √ 2 x 9.8 x .96

= 4.33 m / s