Question

A juggler throws balls into air. He throws one when ever the previous one is at its highest point. If he throws $ n $ balls each second, the height to which each ball will rise is

• A. $ \frac{g}{2n^{2}} $
• B. $ \frac{2g}{n^{2}} $
• C. $ \frac{2g}{n} $
• D. $ \frac{g}{4n^{2}} $

Answer 1

Answer: (A)

Time taken by each ball to reach highest point, $ t=\frac{1}{n} $ second.
As the juggler throws the second ball, when the first ball is at its highest point, so $ v = 0 $
Using $ v = u + at $ , we have
$ 0 = u + (- g) (1 /n) $ or $ u = (g/n) $ . Also, $ v^2 = u^2 + 2as $
$ \therefore 0=\left(g/n^{2}\right)+2\left(-g\right)h $ or $ h=\frac{g}{2n^{2}} $