A ball is dropped on to a fixed horizontal surface from a height 'h'. The coefficient of restitution is 'e'. The average velocity of the ball from the instant it is dropped till it goes to maximum height after first impact with ground is
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a
(1+e1−e)2gh
b
(1+e21+e)gh2
c
(1−e1+e)2gh
d
(1−e)gh2
answer is D.
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Detailed Solution
h1=e2h v=total displacementtotal time =h−h12hg +e2hg=(1−e2)h(1+e)2hg =(1−e)(1+e)(1+e)gh2=(1−e)gh2