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A juggler keeps (n) balls going with one...

A juggler keeps (n) balls going with one hand so that at any instant, ( n-1) balls in air and one ball in the hand, If each ball rises to a height of (x) meter, find the time for which each ball stays the hand.

A

`1/n -1((sqrt 2x)/g)`

B

`2/n - 1((sqrt 2x)/g)`

C

`2/n((sqrt 2x)/ g)`

D

`1/n((sqrt 2x)/ g)`

Text Solution

Verified by Experts

The correct Answer is:
(b)
Let u be the initial velocity of the ball while going upwards. The final velocity of the ball at height x is, v = 0.
So u = `sqrt(2gx)`
Time of flight, `T = (2u)/g = 2/g sqrt(2gx) = 2 sqrt(2x)/g`
During time T, (n -1) balls will be in air and one ball will be in hand. So time for one ball in hand
`T /(n - 1) = 2 sqrt(2x)/g(n - 1) = 2 / n - 1 ((sqrt 2x)/g)`
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