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Question

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

h1h0=v21v20;

h2h1=v22v21;
hnhn1=v2nv2n1;
; hnh0=e2n;

e2=64cm1m=0.64=0.82

; hnh0=(0.8)2n

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