Q. A juggler throws balls into air. He throws one whenever the previous one is at its highest point. If the 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) $\frac{g}{4n^{2}}$
Explanation
(a) Time taken by each ball to reach highest point,
$t = \frac{1}{n}$ s.
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 $v^{2} = u^{2} + 2as$
$\therefore$ $0 = (g/n)^{2} + 2( - g)h$ or $h = \frac{g}{2n^{2}}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.