If the ball hits the $n^{th}$ 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 displacements. Velocity along horizontal direction $= u$ (remains constant) and initial vertical velocity $= zero.$

$\therefore nb=ut and nh= 0+\frac{1}{2}g{t}^2$

Eliminating $t$ from the equation

$nh=\frac{1}{2}g{\left(\frac{nb}{u}\right)}^2 \Rightarrow n=\frac{2h{u}^2}{g{b}^2}$