A ball rolls off to the top of a staircase with a horizontal velocity  $u\hspace{0.17em}\hspace{0.17em}m/s.$ If the steps are $h$ metre high and $b$ metre wide, the ball will hit the edge of the $n^{th}$ step, if

If the ball hits the nth step, then the horizontal distance traversed $=nb.$ 

Here, the velocity along the horizontal direction $=u.$   

Intitial velocity along the vertical direction  $=0.$  

So  $nb=ut$                  (i)
             

 $nh=0+\frac{1}{2}g{t}^2$                            (ii)
From $t=\frac{nb}{v},$  

putting $t$  in equation (ii)
 $nh=\frac{1}{2}g\times {\left(\frac{nb}{u}\right)}^2$or  $n=\frac{2h{u}^2}{g{b}^2}$