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A boy throws n balls per second at regul...

A boy throws n balls per second at regular time intervals. When the first ball reaches the maximum height, he throws the second one vertically up. The maximum height reached by each ball is

A

`(g)/(2(n-1)^(2))`

B

`(g)/(2n^(2))`

C

`(g)/(n^(2))`

D

`(g)/(n)`

Text Solution

Verified by Experts

The correct Answer is:
B
Time interval between two balls = Time of ascend
` = (1)/(n) = (u)/(g) u = (g)/(n), " " h = (u^(2))/(2g)`
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