A ball is is dropped on the floor from a height of 10m. It rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01s, the average acceleration during contact is
A
2100m/s2 downwards
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B
2100m/s2 upwards
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C
1400m/s2
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D
700m/s2
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Solution
The correct option is B2100m/s2 upwards Velocity of ball at the time of striking the floor, u=√(2gh1)=√(2×9.8×10) =14m/s (downwards) Velocity with which it rebounds, v=√(2gh2)=√(2×9.8×2.5) =7m/s (upwards) Average Acceleration during contact is a=v−ut2−t1=7−(−14)0.01 =210.01=2100m/s2 upwards (Taking upward as positive and downward as negative )