Q. A ball rolls of the top of a stair way with a horizontal velocity $u\ ms^{- 1}$. If the steps are $h$ metre and $b$ metre wide, the ball hits the edge of $n$th step, the time taken by the ball is

A) $\frac{hu}{gb}$
B) $\frac{2hu}{gb}$
C) $\frac{2hu^{2}}{gb}$
D) $\frac{hu^{2}}{2gb}$
Explanation

(b) The ball reaches $n$th step in time $t$, then $bn = ut$
or $t = bn/u$,
$nh = \frac{1}{2}gt^{2} = \frac{1}{2}g \times \frac{b^{2}n^{2}}{u^{2}}$; so $n = \frac{2u^{2}h}{gb^{2}}$
Time taken to travel vertical distance $nh$ is
$t = \sqrt{\frac{2nh}{g}} = \sqrt{\frac{2h}{g} \times \frac{2u^{2}h}{gb^{2}}} = \frac{2uh}{gb}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.