A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, the average acceleration during contact is

Velocity at the time of striking the floor,u=2gh1=2×9.8×10=14m s-1Velocity with which it rebounds.v=2gh2=2×9.8×2.5=7m s-1∴  Change in velocityΔv=7−(−14)=21ms−1Acceleration=ΔvΔt=21001=2100ms−2 (upwards)