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Q.

A ball is projected vertically down with an initial velocity from a height of 20m on to a horizontal floor. During the impact it loses 50% of its energy and rebounds to the same height. The velocity of projection is (g = 9.8ms−2 )

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a

20ms−1

b

15ms−1

c

10 ms−1

d

5 ms−1

answer is A.

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Detailed Solution

50100(mgh+12mV2)=mgh

12(10×20+12V2)=10×20

V22=200⇒v=400=20ms−1

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