Correct Answer - C
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 +(1//2) g t^(2)` .......(ii)
From Eqs. (i) and (ii), we get by eliminating t,
`nh = (1//2) g (nb//u)^(2) rArr u = (2hu^(2))/(gb^(2))`