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=14 m/s
Velocity with which it rebounds.
v=√2gh2=√2×9.8×2.5=7 m/s
∴ Change in velocity Δv=7−(−14)=21 m/s
Acceleration = ΔvΔt=210.01=2100 m/s2 (upwards)