1. Average velocity for the first projectile

For the first flight:

• Initial speed $= u_0$
• Projection angle $= \theta$

Time of flight: $T_1 = \frac{2u_0\sin\theta}{g}$

Horizontal range: $R_1 = u_0\cos\theta \cdot T_1 = u_0\cos\theta \cdot \frac{2u_0\sin\theta}{g} = \frac{u_0^2\sin 2\theta}{g}$

Since the ball lands back on the ground, net displacement during the first flight is purely horizontal and equals $R_1$.

Hence magnitude of average velocity for first flight is $V_1 = \frac{R_1}{T_1} = \frac{\frac{u_0^2\sin 2\theta}{g}}{\frac{2u_0\sin\theta}{g}} = u_0\cos\theta$

So, $V_1 = u_0\cos\theta$

2. Motion after successive rebounds

After first hit, the ball rebounds with speed $\frac{u_0}{\alpha}$ at the same angle $\theta$.

Then after each subsequent bounce, the same reduction factor continues, so speeds of successive flights are: $u_0,\; \frac{u_0}{\alpha},\; \frac{u_0}{\alpha^2},\; \frac{u_0}{\alpha^3},\dots$

For the $n$-th flight with speed $u$, we have:

• Time of flight proportional to $u$
• Range proportional to $u^2$

Thus total time of motion is $T_{\text{tot}} = \frac{2\sin\theta}{g}\left(u_0 + \frac{u_0}{\alpha} + \frac{u_0}{\alpha^2}+\cdots\right)$ $T_{\text{tot}} = \frac{2u_0\sin\theta}{g}\left(1 + \frac{1}{\alpha} + \frac{1}{\alpha^2}+\cdots\right)$

This is a GP, so for $\alpha>1$, $\sum_{n=0}^{\infty} \frac{1}{\alpha^n} = \frac{1}{1-1/\alpha} = \frac{\alpha}{\alpha-1}$

Hence, $T_{\text{tot}} = \frac{2u_0\sin\theta}{g}\cdot \frac{\alpha}{\alpha-1}$

3. Total horizontal displacement

Total displacement is purely horizontal, equal to sum of all ranges: $R_{\text{tot}} = \frac{\sin 2\theta}{g}\left(u_0^2 + \frac{u_0^2}{\alpha^2} + \frac{u_0^2}{\alpha^4}+\cdots\right)$ $R_{\text{tot}} = \frac{u_0^2\sin 2\theta}{g}\left(1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4}+\cdots\right)$

Again this is a GP: $\sum_{n=0}^{\infty} \frac{1}{\alpha^{2n}} = \frac{1}{1-1/\alpha^2} = \frac{\alpha^2}{\alpha^2-1}$

So, $R_{\text{tot}} = \frac{u_0^2\sin 2\theta}{g}\cdot \frac{\alpha^2}{\alpha^2-1}$

4. Magnitude of average velocity for entire motion

$V_{\text{avg}} = \frac{R_{\text{tot}}}{T_{\text{tot}}}$

Substitute: $V_{\text{avg}} = \frac{\frac{u_0^2\sin 2\theta}{g}\cdot \frac{\alpha^2}{\alpha^2-1}}{\frac{2u_0\sin\theta}{g}\cdot \frac{\alpha}{\alpha-1}}$

Use $\sin 2\theta = 2\sin\theta\cos\theta$: $V_{\text{avg}} = \frac{u_0^2(2\sin\theta\cos\theta)}{g}\cdot \frac{\alpha^2}{\alpha^2-1} \cdot \frac{g}{2u_0\sin\theta} \cdot \frac{\alpha-1}{\alpha}$

Simplifying, $V_{\text{avg}} = u_0\cos\theta \cdot \frac{\alpha^2}{\alpha^2-1}\cdot \frac{\alpha-1}{\alpha}$

Now, $\alpha^2-1=(\alpha-1)(\alpha+1)$

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

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

Given, $V_{\text{avg}} = 0.8V_1$

Thus, $\frac{\alpha}{\alpha+1} = 0.8 = \frac{4}{5}$

Solve: $5\alpha = 4(\alpha+1)$ $5\alpha = 4\alpha + 4$ $\alpha = 4$

5. Final answer

$\boxed{4}$