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

Answer: 4

$V_{\text{avg }}=\frac{R_1+R_2+R_3+\dots \dots }{T_1+T_2+T_3+\dots \dots }$
$V_{avg}=\frac{\frac{2u_{x_1}{u}_{y_1}}{g}+\frac{2u_{x_2}{u}_{y_2}}{g}+\dots .}{\frac{2u_{y_1}}{g}+\frac{2u_{y_2}}{g}+\dots ..}$
$=\frac{u_{0}cos\theta \cdot u_{0}sin\theta +\frac{u_0}{\alpha }cos\theta \cdot \frac{u_0}{\alpha }sin\theta +\dots .}{u_{0}sin\theta +\frac{u_0}{\alpha }sin\theta +\frac{u_0}{{\alpha }^2}sin\theta +\dots .}$
$=\frac{u_{0}cos\theta \left[1+\frac{1}{{\alpha }^2}+\frac{1}{{\alpha }^4}+\dots \cdot \frac{1}{{\alpha }^{2n}}\right]}{\left[1+\frac{1}{\alpha }+\frac{1}{{\alpha }^2}+\dots \dots +\frac{1}{{\alpha }^n}\right]}$
Given, $u_{0}cos\theta =V_1$
$\&V_{avg}=0.8v_1$
$0.8v_1=\frac{v_1\left[1+\frac{1}{{\alpha }^2}+\frac{1}{{\alpha }^4}+\dots ..+\frac{1}{{\alpha }^{2n}}\right]}{1+\frac{1}{\alpha }+\frac{1}{{\alpha }^2}+\dots .+\frac{1}{{\alpha }^n}}$
$0.8=\frac{\left[\frac{1}{1-\frac{1}{{\alpha }^2}}\right]}{\left[\frac{1}{1-\frac{1}{\alpha }}\right]}$
$=\frac{\frac{1}{\left(1-\frac{1}{\alpha }\right)\left(1+\frac{1}{\alpha }\right)}}{\frac{1}{\left(1-\frac{1}{\alpha }\right)}}$
$0.8=\frac{1}{\left(1+\frac{1}{\alpha }\right)}$
$1+\frac{1}{\alpha }=\frac{1}{0.8}$
$\frac{1}{\alpha }=\frac{10}{8}-1$
$\frac{1}{\alpha }=\frac{2}{8}$
$\alpha =\frac{8}{2}=4$