A ball is thrown from ground at an angle θ with horizontal and with an initial speed u 0 . For the resulting…

A ball is thrown from ground at an angle θ with horizontal and with an initial speed u0. 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 V1. After hitting the ground, ball rebounds at the same angle θ but with a reduced speed of u0α. 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 V1, the value of α is______

Solution

For the first projectile, the average velocity is given by $V = \frac{R}{T} = U_x = V_1$, where $R$ is the range and $T$ is the time of flight. The range and time are given by $\begin{aligned} R &= \frac{2U_xU_y}{g}, \\ T &= \frac{2U_y}{g}. \end{aligned}$ For the journey, the average velocity is given by $V_{1 \rightarrow n}$ = $\frac{R_1 + R_2 + \ldots + R_n}{T_1 + T_2 + \ldots + T_n}$ = $\frac{\frac{2u_{x1}u_{y1}}{g} + \frac{2u_{x2}u_{y2}}{g} + \ldots + \frac{2u_{xn}u_{yn}}{g}}{\frac{2u_{y1}}{g} + \frac{2u_{y2}}{g} + \ldots + \frac{2u_{yn}}{g}}$. This implies that $U_x \left[ \frac{1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \ldots + \frac{1}{\alpha^{2n}}}{1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} + \ldots + \frac{1}{\alpha^n}} \right] = 0.8v_1, where 1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \ldots + \frac{1}{\alpha^{2n}}$ implies a geometric progression and the sum of the geometric progression is given by $\frac{1}{1 - \alpha^2}$. Therefore, $\frac{V_1 \left[ \frac{1}{1 - \frac{1}{\alpha^2}} \right]}{\left[ \frac{1}{1 - \frac{1}{\alpha}} \right]} = 0.8v_1 implies that \frac{\alpha}{1 + \alpha} = 0.8, which gives \alpha = 4$.

Asked in: JEE Advanced 2019 (Paper 2)

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