Answer:
Let
u be the initial velocity of the ball while going upwards. The final velocity
of the ball at height x is, \[\upsilon =\text{ }0\].
Using the relation; \[{{\upsilon
}^{\text{2}}}={{\text{u}}^{\text{2}}}+\text{2as}\], we have
\[0={{\text{u}}^{\text{2}}}-\text{2gx}\]or \[\text{u}=\sqrt{2gx}\]
If t is the time taken by the ball in going up through distances,
then\[~0\text{ }=\text{u }+(-\text{g})\text{ t}\] or\[\text{t }=\frac{u}{g}\]. Total
time after which the ball comes into the hand is
\[\text{T}=\text{2t}=\frac{2u}{g}=\frac{2}{g}\sqrt{2gx}=2\sqrt{\frac{2x}{g}}\]
During time T, (n - 1) balls will be in air and one ball will be
in hand.
Time for one ball in hand =\[\frac{T}{n-1}=\frac{2\sqrt{2x/g}}{(n-1)}=\frac{2}{(n-1)}\sqrt{\frac{2x}{g}}\].
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