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Question

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

A
2100 m/sec2 (downwards)
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B
2100 m/sec2 (upwards)
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C
1400 m/sec2
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D
700 m/sec2
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Solution

The correct option is B 2100 m/sec2 (upwards)

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)


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