A ball rolls off top of a stair way with a horizontal velocity μ ms-1. If the steps are h metres high and b metres wide, the ball will just hit the edge of nth step, if n equals to

If the ball hits the nth step, the horizontal and vertical distances traversed are nb and nh, respectively. Let t be the time taken by the ball for these horizontal and vertical displacement. Then, velocity along horizontal direction remains constant = u, initial vertical velocity is zero.∴ nb = ut                       … (i)     nh = 0 + (l/2) gt2           …(ii)From Eqs. (i) and (ii), we get by eliminating t. nh=(1/2)g(nb/u)2⇒n=2hu2gb2