Is it possible to calculate the time in which a bouncing ball with restitution coefficient $k$ which first hits the ball with velocity $v$ will be considered stopped? Note that each collision lasts $\tau$ seconds.
I know how to calculate the answer if $\tau \to 0$. Let's see! Let's denote $t = \frac{kv}{g}$ the time after the ball fill first reach velocity $0$ at a height $h$. So, between the first collisions, we have a total time $t_1 = 2t = \frac{2kv}{g}$. We can prove easily (I think we can even try mathematical inductory reasoning) that $t_n = \frac{2k^nv}{g}$. Let's compute the infinite sum/series: $$T = \sum_{i = 1}^\infty t_i = \frac{2v}{g} \sum_{i = 1}^\infty k^n$$
As $k < 1$ (restitution coefficient), we can enter Geometric series: $$T = \frac{2v}{g} \cdot \frac{k}{1 - k}$$
Howerevr, if $\tau \gg 0$, the times are $t_n = \frac{2k^nv}{g} + \tau$. So the series is: $$T = \frac{2v}{g} \sum_{i = 1}^\infty k^n + \sum_{i = 1}^\infty \tau$$
Calculating the Geometric series and the divergent sum: $$T = \frac{2v}{g} \cdot \frac{k}{1 - k} + \infty = \infty$$
So, the time looks infinite... Is this true? Just by taking into account a collision time/delay, the process should be considered infinite (in time)? Or, where are my mistakes and what should I do to avoid this situation? Thanks a lot!
