A ball is dropped from rest at a height $h_0$ and bounces from a surface such that the height of the $n$th bounce, $h_n$ is given by $h_n=αh_{n-1}$, where $h_{n-1}$ is the height of the previous, $(n-1)$th bounce. The factor $α$ has value between 0 and 1.
How long does the ball take to cover the distance it travels before coming to rest?
I calculated the total distance travelled by the ball before coming to rest to be $\frac{2h_0}{1-α}-h_0$, using the infinite sum of $x^n$, where $|x|<1$, which is equal to $\frac{1}{1-x}$. Note that after the first bounce, the distance of every subsequent "bounce" is doubled, as the ball falls to the ground then bounces up to the same height again.
In order to calculate how long the ball takes to cover this distance, I must not assume an average velocity. Therefore, I thought to use the SUVAT equation $s=ut+\frac{1}{2}at^2$, where $a=g=9.81ms^{-2}$, to find the time taken for each bounce. However, the question hints towards the solution involving the same infinite sum used as above to calculate the total distance, so I'm a bit lost as to how I can apply this.
The question can’t be answered in its current form because the answer depends on when and where the energy is lost. The ball loses energy both from the inelastic bounce on the ground and from air friction. It would be quite complicated to solve this taking both effects into account simultaneously, so I suspect that the intended interpretation is that one of the effects is to be neglected. Since the equation you want to use assumes that there’s no air friction, I’ll make that assumption (so we’re talking e.g. about a basketball and not a table tennis ball).
It’s convenient to make the motion symmetric about $t=0$ by placing the origin at the time when the ball is at its peak (and thus at rest). Then the downward motion is described by $s=\frac12gt^2$, and setting $s=h_n$ yields $\frac12gt_n^2=h_n$ and thus $t_n=\sqrt{\frac{2h_n}g}$. The time the ball takes to rise and fall is twice this time. The recurrence relation for the times is therefore $t_n=\sqrt{\alpha}t_{n-1}$. I’ll leave the remaining details to you.