A ball is thrown horizontally from a height h above a staircase as shown in the figure. If the coefficient…

Vertical velocity of the ball just before each collision is

$v_{\mathrm{app}}=\sqrt{2gh}$

The height up to which the ball bounces after each collision is

$h^{\text{'}}=h-d$

Vertical velocity of the ball just after the collision is

$v_{\mathrm{sep}}=e{v}_{\mathrm{app}}=e\sqrt{2gh}$

Using the principle of conservation of mechanical energy we get

$\frac{1}{2}m{\left(e\sqrt{2gh}\right)}^2=mg\left(h-d\right)$

$\frac{1}{2}{e}^2\left(2gh\right)=g\left(h-d\right)$

$e^{2}h=h-d$

$d=h\left(1-e^2\right)$

$h=\frac{d}{1-e^2}$