A ball is thrown from ground at an angle $θ$ 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 $θ$ but with a reduced speed of $\frac{u_{0}}{α}$ . 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 $α$ is……..

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answer is 4.

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Detailed Solution

$T_{0} = \frac{2 μ_{0} \sin θ}{g} V_{a v g} = (μ_{0} \cos θ) T_{0} + (\frac{μ_{0}}{α} \cos θ) \frac{T_{0}}{α} + ....... \frac{}{T_{0} + \frac{T_{0}}{α} + \frac{T_{0}}{α^{2}} + .....} \frac{= (μ_{0} \cos θ) T_{0} [1 + \frac{1}{α^{2}} + \frac{1}{α^{4}} +]}{T_{0} (1 + \frac{1}{α} + \frac{1}{α^{2}} ....)} V_{a v g} = V_{1} (\frac{α}{1 + α}) 0.8 V_{1} = V_{1} (\frac{α}{1 + α}) α = 4$