A ball is thrown from ground at 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 $u_{0}/α.$ 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 ____

Step 1: Analyze the first projectile motion.

The horizontal range $R_1$ and the time of flight $T_1$ for the first projectile motion are given by the standard formulas.

$$R_1 = \frac{u_0^2 \sin 2\theta}{g}$$ $$T_1 = \frac{2u_0 \sin\theta}{g}$$

Step 2: Calculate the average velocity for the first flight.

The magnitude of the average velocity $V_1$ is the ratio of the horizontal displacement to the time of flight.

$$V_1 = \frac{R_1}{T_1}$$ $$V_1 = \frac{\frac{u_0^2 \sin 2\theta}{g}}{\frac{2u_0 \sin\theta}{g}}$$ $$V_1 = u_0 \cos\theta$$

Step 3: Calculate the total horizontal displacement.

Since the speed is reduced by a factor of $\alpha$ after each bounce, the range of successive bounces decreases by a factor of $\alpha^2$.

$$X = \sum_{n=1}^{\infty} R_n = R_1 + \frac{R_1}{\alpha^2} + \frac{R_1}{\alpha^4} + \dots$$ $$X = \frac{R_1}{1 - \frac{1}{\alpha^2}}$$ $$X = \frac{u_0^2 \sin 2\theta}{g} \cdot \frac{\alpha^2}{\alpha^2 - 1}$$

Step 4: Calculate the total time of flight.

The time of flight for successive bounces decreases by a factor of $\alpha$.

$$T = \sum_{n=1}^{\infty} T_n = T_1 + \frac{T_1}{\alpha} + \frac{T_1}{\alpha^2} + \dots$$ $$T = \frac{T_1}{1 - \frac{1}{\alpha}}$$ $$T = \frac{2u_0 \sin\theta}{g} \cdot \frac{\alpha}{\alpha - 1}$$

Step 5: Determine the average velocity for the entire motion.

The average velocity for the entire duration is the ratio of the total displacement to the total time.

$$V_{\text{avg}} = \frac{X}{T}$$ $$V_{\text{avg}} = \frac{\frac{u_0^2 \sin 2\theta}{g} \cdot \frac{\alpha^2}{\alpha^2 - 1}}{\frac{2u_0 \sin\theta}{g} \cdot \frac{\alpha}{\alpha - 1}}$$ $$V_{\text{avg}} = u_0 \cos\theta \cdot \frac{\alpha}{\alpha + 1}$$

Step 6: Solve for $\alpha$.

Equate the total average velocity to $0.8\,V_1$.

$$V_{\text{avg}} = 0.8\,V_1$$ $$u_0 \cos\theta \cdot \frac{\alpha}{\alpha + 1} = 0.8\,u_0 \cos\theta$$ $$\frac{\alpha}{\alpha + 1} = 0.8$$ $$\alpha = 4$$

Answer: 4