Infinitely bouncing ball

JEE Advanced 2019 Paper 2, Question 9

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 0.8 V_{1}, what is the value of \alpha?

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The average velocity of the ball between bounces is just the horizontal velocity. Remember, velocity is displacement over time and the net displacement between successive bounces is just the distance covered on the horizontal surface. (The wording ‘average velocity’ confused me when I first read this problem, hence the clarification.)

We also know that the horizontal component of velocity is a constant for projectile motion; up to the first bounce it is just u_0 \cos \theta. So, V_1 = u_0 \cos \theta. Furthermore the ball spends a time t_1 in the air before the first bounce, where

(1)   \begin{equation*}   u_0 \sin \theta = \frac{g t_1}{2} \implies t_1 = \frac{2 u_0 \sin \theta}{g} . \end{equation*}

The first equation follows from the fact that the vertical velocity u_0 \sin \theta is deccelerated to 0 by gravity in the time t_1/2 that it takes to reach the peak of the projectile.

Similarly, the average veloctity between the first and second bounce is V_2 = \frac{u_0}{\alpha} \cos \theta \equiv V_1/\alpha and the ball travels for a time t_2 = \frac{2 u_0 \sin \theta}{\alpha g} = t_1/\alpha.
Proceeding in this manner we find that the average velocity between the (m-1)^{\rm th} bounce and the m^{\rm th} one is V_m = V_1/\alpha^{m-1}, and the time of travel is t_m = t_1/\alpha^{m-1}. To find the average velocity, we find the total distance covered by the total time,

(2)   \begin{align*}   V_{\rm avg}     &= \frac{V_1 t_1 + V_2 t_2 + V_3 t_3 + \dots}{t_1 + t_2 + t_3 + \dots} \nonumber \\     &= \frac{V_1 t_1 \left( 1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \dots \right)}{t_1 \left( 1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} + \dots \right)} \nonumber \\     &= \frac{V_1 \left( \frac{1}{1-\frac{1}{\alpha^2}} \right)}{\frac{1}{1-\frac{1}{\alpha}}}     = \frac{V_1}{1+\frac{1}{\alpha}} . \end{align*}

where, in the second to last equality, we’ve used the formula for the sum of an infinite geometric series: 1+r+r^2+r^3+\dots = \frac{1}{1-r}, which converges for r<1. We’re told that V_{\rm avg} = 0.8 V_1, and therefore

(3)   \begin{equation*}   \frac{1}{1+\frac{1}{\alpha}} = \frac{4}{5}   \implies   \boxed{\alpha = 4} . \end{equation*}

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