Answer
a)
$D=\frac{H(1+r)}{1-r}$
b)
$\sqrt (\frac{2H}{g}) \times \frac{1+\sqrt r}{1-\sqrt r}$
c)
$T=\sqrt{\frac{2H}{g}} \times \frac{1+k}{1-k}$
Work Step by Step
a)
The ball falls an initial height of $H$ meters then climbs a height of $rH$ meters.
Then falls to a height of $rH$ meters and climbs a height of $r^{2}H$ meters.
Then falls to a height of $r^{2}H$ meters and then climbs a height of $r^{3}H$ meters.
Then falls to a height of $r^{3}H$ meters and so on...
$D=H+rH + rH +r^{2}H +r^{2}H +r^{3}H +r^{3}H+...$
$D=H + 2rH +2r^{2}H +2r^{3}H...$
$D=-H + 2H +2rH +2r^{2}H+2r^{3}H...$
$D= [\Sigma^{\infty}_{n=0}2Hr^{n}] - H$
$D=[2H\Sigma^{\infty}_{n=0}r^{n}] - H$
$D=[2H\frac{1}{1-r}] - H$
$D=\frac{2H}{1-r}-H$
$D=\frac{2H-H(1-r)}{1-r}$
$D=\frac{2H-H+Hr}{1-r}$
$D=\frac{H+Hr}{1-r}$
$D=\frac{H(1+r)}{1-r}$
b)
$\frac{1}{2}gt^{2}_{n}=Hr^{n}$
$t^{2}_{n}=\frac{2H}{g}r^{n}$
$t_{n} = \sqrt (\frac{2H}{g})(\sqrt r)^{n}$
$T=t_{0} +2t_{1} +t^{2}_{2}+...=\Sigma^{\infty}_{n=0}2\sqrt (\frac{2H}{g})(\sqrt r)^{n}-t_{0}$
$=2\sqrt (\frac{2H}{g})\Sigma^{\infty}_{n=0}2(\sqrt r)^{n}-\sqrt (\frac{2H}{g})$
$=\sqrt (\frac{2H}{g})(\frac{2}{1-\sqrt r}-1)$
$=\sqrt (\frac{2H}{g}) \times \frac{1+\sqrt r}{1-\sqrt r}$
c)
Calculate the time it takes to decelerate down from $k^{n}v$ meters per second to $0$, or equivalently accelerate from $0$ to $k^{n}v$.
$gt_{n}=k^{n}v$
$t_{n} = \frac{k^{n}v}{g}$
$T= t_{0} +2t_{1} +2t_{2} +.... = \Sigma^{\infty}_{n=0} \frac{2v}{g}k^{n} - \frac{v}{g}$
$=\frac{v}{g}(\Sigma^{\infty}_{n=0}2k^{n}-1)$
$=\frac{v}{g}(\frac{2}{1-k}-1)$
$=\frac{v}{g} \times \frac{1+k}{1-k}$
We also know from physics that:
$v_f^2=v_i^2+2gH$
Since the ball is dropped with initial velocity 0, $v_i=0$:
$v_f^2=0+2gH$
$v=\sqrt{2gH}$
Therefore:
$T=\frac{\sqrt{2gH}}{g} \times \frac{1+k}{1-k}$
$T=\sqrt{\frac{2H}{g}} \times \frac{1+k}{1-k}$
