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, ball rebounds at the same angle $\theta$ but with a reduced speed of $\frac{u_0}{\alpha }$. Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for the entire duration of motion is  $0.8v_1$, the value of $\alpha$ is_________.

Average velocity of entire motion for long time as specified in the question is given as
Average Velocity  $=\frac{Totaldisplacement}{Totaltime}$
 Total time $=t_1+t_2+t_3+\mathrm{...}=t_1+\frac{t_1}{\alpha }+\frac{t_1}{{\alpha }^2}+\mathrm{...} \Rightarrow$ Total time $=\frac{t_1}{1-\frac{1}{\alpha }}$  
If average velocity for the first projectile is $v_1$ then it is equal to  $u\cos \theta$, thus for the second projectile it will be $v_1/\alpha$, for the third projectile it will be $v_1/{\alpha }^2$ and so on. Thus total displacement is given as
 Total displacement $=v_1{t}_1+\frac{v_1}{\alpha }\cdot \frac{t_1}{\alpha }+\mathrm{...}=\frac{v_1{t}_1}{1-\frac{1}{{\alpha }^2}}$
Solving for average velocity, it gives $<v>=\frac{v_1\alpha }{\alpha +1}=0.8v_1 \Rightarrow \alpha =4$