A ball initially at rest falls from a height h=2.5m. After collision with surface having value of coefficient of restitution e=0.6, it rebounds back. Find the rebound velocity of ball.
A
2m/s
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B
4.2m/s
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C
6m/s
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D
11.66m/s
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Solution
The correct option is B4.2m/s Let the velocity of ball just before collision is u1m/s. We know that, For free fall, u1=√2gh ∴u1=√2×9.8×2.5=7m/s Let the velocity of ball after collision be v1m/s. We know that, e=v2−v1u1−u2 Initial and final velocity of surface is zero ∴u2&v2 are equal to zero. ⇒e=0−v17−0 ⇒v1=−e×7 ⇒v1=−0.6×7 ⇒v1=−4.2m/s So, the rebound velocity of the ball is 4.2m/s.