A steel ball is dropped on a hard surface from a height of 1 m and rebounds to a height of 64 cm. The maximum height attained by the ball after nth bounce is (in m)-
A
(0.64)2n
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B
(0.8)2n
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C
(0.5)2n
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D
(0.8)n
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Solution
The correct option is A(0.8)2n Let vn= velocity after nth bounce hn= height after nth bounce h0=1m=100 cm h1=64 cm e= velocity of separation/velocity of approach=h1h0 v20=2gh0 v21=2gh1