(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
$t = \sqrt{\frac{2nh}{g}} = \sqrt{\frac{2h}{g} \times \frac{2u^{2}h}{gb^{2}}} = \frac{2uh}{gb}$