Problem source

A ball of mass $m$ is thrown vertically upwards from the floor of a room of height $h$ with speed $\sqrt{2kgh},$ where $k>1.$ The coefficient of restitution between the ball and the ceiling or floor is $a$. Both the ceiling and floor are level. Show that the kinetic energy of the ball immediately before hitting the ceiling for the $n$th time is 
\[
mgh\left(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1}\right).
\]
Hence show that the number of times the ball hits the ceiling is at most 
\[
1-\frac{\ln[a^{2}(k-1)+k]}{4\ln a}.
\]

Solution source

\begin{align*}
&& \text{energy when projected}  &= \frac12 m(2kgh) \\
&&&= kghm \\
&& \text{energy when hitting ceiling the first time}  &= mgh + \frac12 m v^2 \\
\text{COE}: && kghm &= mgh + \frac12 mv^2 \\
\Rightarrow && v^2 &= 2gh(k-1)
\end{align*}
It will rebound with speed $\sqrt{2gh(k-1)}a$.
\begin{align*}
&& \text{energy when rebounding from ceiling} &=gh(k-1)a^2 + mgh \\
&& \text{energy before hitting the floor} &= \frac12 mv^2 \\
\text{COE}: && gh(k-1)a^2 + mgh &= \frac12 mv^2 \\
\Rightarrow && v^2 &= 2gh((k-1)a^2+1)
\end{align*}

The ball will rebound with kinetic energy $m gh((k-1)a^2+1)a^2 = mgh((k-1)a^4+a^2)$

And will reach the ceiling with kinetic energy $mgh((k-1)a^4+a^2-1)$.

When $n = 1$, the kinetic energy (before hitting the ceiling for the first time) is $mgh(k-1)$. Suppose $s_n$ is the expression for the kinetic energy divided by $mgh$, ie $s_1 = k-1$, then:

Clearly $s_1 = k-1 = a^{4\cdot1-4}(k-1) + \frac{a^{4\cdot-4}-1}{a^2+1}$, so our hypothesis holds for $n=1$.

Suppose it is true for $n$, then the $n+1$th time it will be:

\begin{align*}
s_{n+1} &=  s_n a^4+a^2-1 \\
 &=  \left ( a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \right) a^4 + a^2 - 1 \\
&= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4}{a^2+1} + \frac{a^4-1}{a^2+1} \\
&= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-a^4+a^4-1}{a^2+1} \\
&= a^{4(n+1)-4}(k-1) + \frac{a^{4(n+1)-4}-1}{a^2+1}
\end{align*}

Which is our desired expression, therefore it is true by induction.

We wont reach the ceiling if this energy is not positive, ie:

\begin{align*}
&& 0 &\leq a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^{2}+1} \\
\Rightarrow && \frac{1}{a^2+1}&\geq a^{4n-4}\left (k - 1 + \frac{1}{a^2+1} \right) \\
\Rightarrow && a^{4n-4} &\geq \frac{1}{a^2+1} \cdot \frac{1}{k - 1 + \frac{1}{a^2+1}} \\
\Rightarrow && a^{4n-4} &\geq \frac{1}{(k-1)(a^2+1)+1} \\
\Rightarrow && 4(n-1) \ln a &\geq - \ln[(k-1)(a^2+1)+1] \\
\underbrace{\Rightarrow}_{\ln a < 0} && (n-1) &\leq \frac{ - \ln[(k-1)(a^2+1)+1]}{4\ln a} \\
\Rightarrow && n & \leq 1 -\frac{ \ln[(k-1)(a^2+1)+1]}{4\ln a} \\
&&&= 1 -\frac{ \ln[(k-1)a^2+k]}{4\ln a} 
\end{align*}