A ball is projected in a direction inclined to the horizontal and it bounces on a horizontal plane. If the range of first rebound is R and coefficient of restitution is e, then range of the next rebound is
A
R′=eR
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B
R′=e2R
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C
R′=Re
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D
R′=R
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Solution
The correct option is AR′=eR
As we know that, Range =ux×T here, ux is horizontal component of velocity ux=ucosθ uy=usinθ= vertical component of velocity Time of flight, T=2usinθg ⇒T=2uyg Hence, Range R=uxT ⇒R=2uxuyg After rebound, horizontal component will not change but vertical component of velocity will be decreased by e times, therefore, u′y=e×uy Therefore new range, R′=2ux×u′yg R′=2euxuyg=e.2uxuyg=eR