If a rubber ball falls from a height h and rebounds upto the height of h / 2. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

Solution

Physics sir · 4 steps

Energy at the top of each bounce is all potential, so compare heights: h before, h/2 after.

1. Energy before. At the start the ball is at rest at height h, so its total energy is mgh, all potential.
2. Energy after. At the top of the rebound it is again momentarily at rest, now at height h/2. Its energy is mgh/2, so the energy lost is mgh/2.

   % loss = (mgh - mgh/2)/mgh × 100 = 50%

3. Speed at impact. On the way down no energy is lost yet, so mgh = 1/2mv². This gives v = √2gh just before the ball hits the ground.
4. So. The loss is 50% and the speed before impact is √2gh, which is option (a).

The idea to keep. Energy is lost only during the collision, not during free fall. So use v = √2gh for the fall, and the ratio of rebound height to drop height for the fraction of energy kept.

Answer from the NTA final answer key. Solution by PrepWiser.