A ball is thrown horizontally from a height h above a staircase as...

Vertical velocity of the ball just before each collision is

${v}_{app} = \sqrt{2gh}$

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

$h^{'} = h - d$

Vertical velocity of the ball just after the collision is

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

Using the principle of conservation of mechanical energy we get

$\dfrac{1}{2}$ m{ \lleft(e\sqrt{2gh})}^{2} = mg \left(h - d\right)$</p><p>$\dfrac{1}{2} e^{2} \left(2gh\right) =$g \left(h - d\right)$

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

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

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