Question

A juggler keeps n balls going with one hand, so that at any instant, (n-1) balls are in air and one ball in the hand. If each ball rises to a height of x metres, the time for each ball to stay in his hand is

• A. $\frac{1}{n-1}\sqrt{\frac{2x}{g}}$
• B. $\frac{2}{n-1}\sqrt{\frac{2x}{g}}$
• C. $\frac{2}{n}\sqrt{\frac{2x}{g}}$
• D. $\frac{1}{n}\sqrt{\frac{2x}{g}}$

Answer 1

Answer: (B)

$\frac{2}{n-1}\sqrt{\frac{2x}{g}}$

Let $u$ be the initial velocity of the ball while going upwards.
The final velocity of the ball at height $x$ is, $v=0$

Using the relation; $v^2=u^2+2as$

we have,

$0=u^2-2gx⟹u=\sqrt{2gx}$

If $t$ is the time taken by the ball in going up through distance $x,$
then $0=u+(-g)t⟹t=\frac{u}{g}$

​Total time after which the ball comes into the hand is :

$T=2t=\frac{2u}{g}=2\sqrt{\frac{2x}{g}}$

During time $T,(n-1)$ balls will be in air and one ball will be in hand.

TIme of stay for each ball in hand , $=\frac{T}{n-1}=\frac{2\sqrt{\frac{2x}{g}}}{n-1}=\frac{2}{(n-1)}\sqrt{\frac{2x}{g}}$
Hence, the correct option is $\left(B\right)$