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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, 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 C 2100m/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=u22gh (v is zero at maximum height)
u2=2g(2.5)
u=7m/s
Time of impact on the ball =0.01s
avg. acceleration=vfvit
a=7(14)0.01 (since both velocities are in opposite directions.)
a=2100m/s2

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