Motion in a Straight Line Part 1

Velocity of rubber ball when it strikes to the ground, $v_{0}=\sqrt{2\ gh_{0}}$ Using the formula of coefficient of restitution, $e=\frac{\text{velocity after collision}}{\text{velocity before collision}}$ $\Rightarrow \;\; e=\frac{v_{1}-0}{v_{0}-0} \Rightarrow e=\frac{v_{1}}{\sqrt{2 g h_{0}}}$ $\Rightarrow \;\; v_{1}=e \sqrt{2 g h_{0}}$...(i) As, initial height of the ball $=h_{0}$. $\therefore$ The first height of the rebound, $h_{1}=\frac{v_{1}^{2}}{2 g}$ $\Rightarrow \;\; h_{1}=e^{2} h_{0}$ [Using Eq. (i)] The nth height of the ball to the rebound, $h_{n}=\frac{v_{n}^{2}}{2 g} \Rightarrow h_{n}=e^{2 n} h_{0}$ The velocity of the ball after $n$th rebound, $v_{n}=e^{n} v_{0}$ Now, the total distance travelled by the ball after nth rebound is $H=h_{0}+2 h_{1}+2 h_{2}+2 h_{3} \dots$ $H=h_{0}+2 e^{2} h_{0}+2 e^{4} h_{0}+2 e^{6} h_{0} \dots$ $H=h_{0}[1+2 e^{2}(1+e^{2}+e^{4}+e^{6} \dots)]$ Using the formula, $1+e^{2}+e^{4}+\dots=\frac{1}{1-e^{2}}$ $\Rightarrow \;\; H=h_{0}\left[1+2 e^{2}\left(\frac{1}{1-e^{2}}\right)\right]$ $\Rightarrow \;\; H=h_{0}\left(\frac{1+ e^{2}}{1- e^{2}}\right)$ $\Rightarrow \;\; H=5\left(\frac{1+(0.81)}{1-(0.81)}\right) \;\; \left(\because e^{2}=\frac{h_{1}}{h_{0}}=\frac{81}{100}\right)$ $H=47.6\ \mathrm{m}$ Now, the total time taken by the ball to come to rest $T=t_{0}+2 t_{1}+2 t_{2}+\dots$ $T=\sqrt{\frac{2 h_{0}}{g}}+2 \sqrt{\frac{2 h_{1}}{g}}+2 \sqrt{\frac{2 h_{2}}{g}}+\dots$ $T=\sqrt{\frac{2 h_{0}}{g}}[1+2 e+2 e^{2}+\dots]$ $T=\sqrt{\frac{2 h_{0}}{g}}[1+2 e(1+e+e^{2}+\dots)]$ $T=19\ \mathrm{s}$ The average velocity, $v_{\text{avg }}=\frac{H}{T}=\frac{47.6}{19}$ $=2.5\ \text{m/s}$ Hence, the average velocity is $2.5\ \text{m/s}$.