A ball is thrown from ground at an angle $ heta$ with horizontal and with an initial speed $u_0$. F

Question

A ball is thrown from ground at an angle $	heta$ 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 $	heta$ 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 $0.8 V_1$, the value of $\alpha$ is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(a) : Assume $u_1, u_2, u_3, \ldots$. be the horizontal velocity of the projectiles as shown in figure. $ 	ext { Average velocity, }\langle vangle=\frac{	ext { Total displacement }}{	ext { Total time taken }} $ \ Time of flight $T=\frac{2 u \sin 	heta}{g}$ For given value of $	heta$, value of $T$ will change with the value of $u$. $	herefore \quad$ Total time taken $=t_1+t_2+t_3+\ldots .$. $=t_1+\frac{t_1}{\alpha}+\frac{t_1}{\alpha^2}+\ldots \ldots=\frac{t_1}{1-\frac{1}{\alpha}}=\frac{t_1 \alpha}{(\alpha-1)}$ Total displacement $=u_1 t_1+u_2 t_2+u_3 t_3+\ldots .$. $$ \begin{aligned} & =u_1 t_1+\frac{u_1}{\alpha} \cdot \frac{t_1}{\alpha}+\frac{u_1}{\alpha^2} \cdot \frac{t_1}{\alpha^2}+\ldots \ldots \ & =\frac{u_1 t_1}{1-\frac{1}{\alpha^2}}=\frac{u_1 t_1 \alpha^2}{\left(\alpha^2-1ight)} \end{aligned} $$ \ $\begin{aligned} & 	herefore \quad\langle vangle=\frac{u_1 t_1 \alpha^2}{\left(\alpha^2-1ight)} 	imes \frac{(\alpha-1)}{t_1 \alpha}=\frac{u_1 \alpha}{\alpha+1} \ & 	ext { or } \quad 0.8 V_1=\frac{u_1 \alpha}{\alpha+1} ...(i) \ & V_1=\frac{u_1 	imes t_1}{t_1}=u_1 ...(ii)\end{aligned} $
From equation (i) and (ii)
$
\alpha=0.8 \alpha+0.8 	ext { or } \alpha-0.8 \alpha=0.8
$
or $\quad 0.2 \alpha=0.8 \quad$ or $\quad \alpha=\frac{0.8}{0.2}=4$