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A juggler throws balls into air. He thro...

A juggler throws balls into air. He throws one when ever the previous one is at its highest point. If he throws n balls each second, the height to which each ball will rise is

A

`g/2n^(2)`

B

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

C

`(2g)/n`

D

`g/(4n^(2))`

Text Solution

Verified by Experts

The correct Answer is:
(a)
Time taken by each ball to reach highest point, t = 1/n second.
As the juggler throws the second ball, when the first ball is at its highest point, so v = 0
Using `v = u + at`, we have
`0 = u + (-g)(1/n)` or `u = (g/n)`. Also, `v^(2) = u^(2) + 2as`
`therefore 0 = (g.n^(2)) + 2(-g)h` or `h = g/2n^(2)`
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