A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, then average acceleration during contact is
A
2100m/s2
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B
1400m/s2
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C
700m/s2
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D
400m/s2
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Solution
The correct option is C2100m/s2 Velocity of the ball on reaching the ground is v2=u2+2as (u is 0) v2=2g(10) v=14m/s After striking the ground the ball bounces off. Velocity of the ball immediately after striking the ground is v2=u2−2gh (v is zero at maximum height) u2=2g(2.5) u=7m/s Time of impact on the ball =0.01s avg. acceleration=vf−vit a=7−(−14)0.01 (since both velocities are in opposite directions.)