AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

The time and speed of the ball to reach on the ground is, $t_{0}=\sqrt{\frac{2h}{g}}$ and $v_{0}=\sqrt{2gh}$ After the first collision, the speed of the ball is, $v_{1}=e v_{0}=e\sqrt{2gh}$ The time when ball stops the motion is, $v=u+at$ $0=v_{1}-g t_{1}$ $t_{1}=\frac{v_{1}}{g}$ The time taken between first and second collision is $2 t_{1}$ . Similarly the time taken between second and third collision is, $2 t_{3}=\frac{2 v_{2}}{g}$ Thus the total time before it ceaser to rebound is, $T=t_{0}+2 t_{1}+2 t_{2}+\dots$ $T=t_{0}+\frac{2 v_{1}}{g}+\frac{2 v_{2}}{g}+\dots$ $T=t_{0}+\frac{2 e v_{0}}{g}+\frac{2 e^{2} v_{0}}{g}+\dots$ $($ As, $v_{2}=e v_{1}=e^{2} v_{0})$ $T=\sqrt{\frac{2h}{g}} [1+2e(1+e+e^{2}+e^{3}+\dots)]$ $($ As, $v_{0}=\sqrt{\frac{2h}{g}} )$ Solve further, $T=\sqrt{\frac{2h}{g}} \left[1+2e\left(\frac{1}{1-e}\right)\right]$ $T=\sqrt{\frac{2h}{g}} \left[\frac{1+e}{1-e}\right]$ $T=\sqrt{\frac{2(180)}{10}} \left(\frac{1+0.5}{1-0.5}\right)$ $T=18\,\mathrm{s}$ Similarly the total distance in the entire motion of ball is, $H=h_{0}+2h_{1}+2h_{2}+\dots$ $H=h+2e^{2}h+2e^{2}h+\dots$ $H=h\left[1+2e^{2}(1+e^{2}+e^{4}+\dots)\right]$ $H=h\left(\frac{1+e^{2}}{1-e^{2}}\right)$ Substitute values, $H=180\left(\frac{1+\frac{1}{4}}{1-\frac{1}{4}}\right)$ $H=300\,\mathrm{m}$ The average speed is, $|v|=\frac{H}{T}$ $|v|=\frac{300}{18}$ $|v|=\frac{50}{3}\ \text{m/s}$ The average velocity is, $v=\frac{180}{18}$ $v=10\ \text{m/s}$