question_answer

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 metre, find the time for which each ball stays in the hand.

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 =\[\frac{T}{n-1}=\frac{2\sqrt{2x/g}}{(n-1)}=\frac{2}{(n-1)}\sqrt{\frac{2x}{g}}\].