Solution

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; $$v^2 = u^2 + 2 as$$ we have, $$0 = u^2 - 2 g x\implies u=\sqrt{2gx}$$ If $$t$$ is the time taken by the ball in going up through distance $$x,$$ then $$0 = u + (- g ) t \implies t = \dfrac ug$$ ​Total time after which the ball comes into the hand is : $$T=2t = \dfrac{2u}{g}= 2 \sqrt{\dfrac{2x}{g}}$$ 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 , $$= \dfrac{T}{n-1}= \dfrac{2 \sqrt{\frac{2x}{g}}}{n-1}= \dfrac{2}{(n-1)}\sqrt{\dfrac{2x}{g}}$$ Hence, the correct option is $$(B)$$