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
Answer: Option B
:
B
Velocity at the time of striking the floor,
u=√2gh=√2×9.8×10=14m/s (-ve since downwards)
Velocity with which it rebounds.
v=√2gh2=√2×9.8×2.5=7m/s (+ve since upwards)
∴ Change in velocity △v=7−(−14)=21m/s
∴ Acceleration = △v△t=210.01=2100m/s2(upwards since positive)
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:
B
Velocity at the time of striking the floor,
u=√2gh=√2×9.8×10=14m/s (-ve since downwards)
Velocity with which it rebounds.
v=√2gh2=√2×9.8×2.5=7m/s (+ve since upwards)
∴ Change in velocity △v=7−(−14)=21m/s
∴ Acceleration = △v△t=210.01=2100m/s2(upwards since positive)
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