When ball dropped from height $h$, then time taken to reach the ground
$$
t_0=\sqrt{\frac{2 h}{g}} \text { and speed, } v_0=\sqrt{2 g h}
$$

After first collision, its speed will become $v_1=e v_0=e \sqrt{2 g h}$
where, $e=$ coefficient of restitution.
Now, the ball will go up and will take time $t_1$, when it stops
$$
\begin{aligned}
& v=u+a t \\
& \quad 0=v_1-g t_1 \\
& t_1=\frac{v_1}{g}
\end{aligned}
$$

It will come down and take same time $t_1$ before second collision.
So, time taken between first and second collision is $2 t_1$.

Similarly, time taken between second and third collision will be $2 t_3=\frac{2 v_2}{g}$

Total time before it ceaser to rebound
$$
\begin{aligned}
& T=t_0+2 t_1+2 t_2+\ldots \ldots \ldots \\
& T=t_0+\frac{2 v_1}{g}+\frac{2 v_2}{g}+\ldots \ldots \ldots \\
& T=t_0+\frac{2 e v_0}{g}+\frac{2 e^2 v_0}{g}+\ldots \ldots \ldots \\
& \quad\left\{\text { as } v_2=e v_1=e^2 v_0\right\} \\
& T=\sqrt{\frac{2 h}{g}}\left[1+2 e\left(1+e+e^2+e^3+\ldots\right)\right] \\
& \left\{\text { as } v_0=\sqrt{\frac{2 h}{g}}\right\}
\end{aligned}
$$

It forms a GP,
$$
\begin{aligned}
T & =\sqrt{\frac{2 h}{g}}\left[1+2 e\left(\frac{1}{1-e}\right)\right] \\
T & =\sqrt{\frac{2 h}{g}}\left(\frac{1+e}{1-e}\right)
\end{aligned}
$$

Given, $h=180 \mathrm{~m}$ and $e=0.5$
$$
\begin{aligned}
T & =\sqrt{\frac{2 \times 180}{10}}\left(\frac{1+0.5}{1-0.5}\right)
\end{aligned}
$$
$$
T=6 \times 3=18 \mathrm{~s}
$$
Calculation of distance in entire motion initial height $h_0=h$

After first collision $=h_1$
As
$$
\begin{aligned}
& e=\frac{v}{u}=\sqrt{\frac{2 g h_1}{2 g h_0}}=\sqrt{\frac{h_1}{h}} \\
& \quad h_1=e^2 h, \text { similarly } h_2=e^2\left(e^2 h\right) \\
& h_2=e^4 h
\end{aligned}
$$

Total distance,
$$
\begin{aligned}
& H=h_0+2 h_1+2 h_2+\ldots \ldots \\
& H=h+2 e^2 h+2 e^4 h+\ldots \ldots \\
& H=h\left[1+2 e^2\left(1+e^2+e^4+\ldots \ldots\right)\right] \\
& H=h\left[1+2 e^2\left(\frac{1}{1-e^2}\right)\right] \\
& H=h\left(\frac{1+e^2}{1-e^2}\right) \\
& H=180\left(\frac{1+\frac{1}{4}}{1-\frac{1}{4}}\right), e=0.5=\frac{1}{2}
\end{aligned}
$$
$\begin{aligned} & H=180 \times \frac{5}{3} \\ & H=300 \mathrm{~m} \\ & \text { Average speed, } v=\frac{\text { Total distance }}{\text { Time }} \\ & v=\frac{300}{18}=\frac{50}{3} \mathrm{~m} / \mathrm{s} \\ & \text { Average velocity, } v=\frac{\text { Total displacement }}{\text { Time }} \\ & v=\frac{180}{18}=10 \mathrm{~m} / \mathrm{s} \\ & \end{aligned}$