Question

A ball is thrown from ground at an angle $\theta$ with horizontal and with an initial speed $u_0$. For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $V_1$. After hitting the ground, the ball rebounds at the same angle $\theta$ but with a reduced speed of $u_0/\alpha$. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is $0.8V_1$, the value of $\alpha$ is

Answer 1

Average velocity $=$ Total displacement$/$Total time
Total time taken $=$ $t_1+t_2+t_3+.....$

$=t_1+\frac{t_1}{\alpha }+\frac{t_1}{{\alpha }^2}+.....$

Total time $=\frac{t_1}{1-\frac{1}{\alpha }}$
Total displacement $=V_1{t}_1+V_2{t}_2+....$
$=V_1{t}_1+\frac{V_1{t}_1}{{\alpha }^2}+\frac{V_1{t}_1}{{\alpha }^4}....$
$=V_1{t}_1[\frac{1}{{\alpha }^2}+\frac{1}{{\alpha }^4}....]$
$=\frac{V_1{t}_1}{1-\frac{1}{{\alpha }^2}}$
on further solving the equation, the average velocity comes as
$=\frac{V_1{\alpha }_1}{\alpha +1}=0.8V_1$
$\therefore$ $\alpha =4.00$