A ball is dropped from height H on to a horizontal surface. If the...

The time it& takes for the ball to fall from a height is given& by

$t = \sqrt{\dfrac{2H}{g}}$

The height up to which the ball bounces after the $n^{th}$~collision from the floor is

${H}_{n} = e^{2n}H$

$t = \sqrt{\dfrac{2H}{g}} + 2\sqrt{\dfrac{2{H}_{1}}{g}} + 2\sqrt{\dfrac{2{H}_{2}}{g}} + \ldots \infty$
t = \sqrt{\dfrac{2H}{g}} \left(1 + 2e + 2e^{2} + \ldots \ldots . . \infty\right)<br />t $=\sqrt{\dfrac{2H}{g}} \left(\dfrac{1 + e}{1 - e}\right)</p>$