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(C) For a projectile launched with speed $u$ at angle $	heta$,the time of flight is $T = \frac{2u \sin 	heta}{g}$ and the horizontal range is $R = \

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$4$

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(C) For a projectile launched with speed $u$ at angle $	heta$,the time of flight is $T = \frac{2u \sin 	heta}{g}$ and the horizontal range is $R = \frac{u^2 \sin 2	heta}{g}$.The average velocity $V_1$ for the first flight is the displacement divided by time: $V_1 = \frac{R}{T} = \frac{u_0^2 \sin 2	heta / g}{2u_0 \sin 	heta / g} = u_0 \cos 	heta$.For the entire motion,the total displacement $S_{total}$ is the sum of all ranges: $S_{total} = R_1 + R_2 + R_3 + \dots = R_1 (1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \dots) = \frac{R_1}{1 - 1/\alpha^2} = \frac{R_1 \alpha^2}{\alpha^2 - 1}$.The total time $T_{total}$ is the sum of all flight times: $T_{total} = T_1 + T_2 + T_3 + \dots = T_1 (1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} + \dots) = \frac{T_1}{1 - 1/\alpha} = \frac{T_1 \alpha}{\alpha - 1}$.The average velocity for the entire motion is $V_{avg} = \frac{S_{total}}{T_{total}} = \frac{R_1 \alpha^2 / (\alpha^2 - 1)}{T_1 \alpha / (\alpha - 1)} = \frac{R_1}{T_1} \cdot \frac{\alpha^2}{\alpha^2 - 1} \cdot \frac{\alpha - 1}{\alpha} = V_1 \cdot \frac{\alpha}{\alpha + 1}$.Given $V_{avg} = 0.8 V_1$,we have $\frac{\alpha}{\alpha + 1} = 0.8$.$\alpha = 0.8 \alpha + 0.8 \implies 0.2 \alpha = 0.8 \implies \alpha = 4$.

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