For a ball falling from rest at height $H$, the coefficient of restitution $0< e<1$ of this surface wrt the ball is defined as the square root of the ratio of the height of one bounce to that of the previous bounce. Simply put, the ball will rebound to a height of $e^2H$.
If $e = 0.8$, and after 5 bounces, the ball is damaged and the coefficient of restitution is reduced to $g$. Find the possible range of values of $g$ if the total distance traveled is more than 4H.
I am quite confused as what the question mean by total distance travelled.
Nonetheless, here are the values I got for some key ideas.
We know that if the coefficient of restitution is $e$, then the distance travelled by the ball just before the $n$ th bounce is $$Sn = (\dfrac{1+e^2-2e^{2n}}{1-e^2})H$$
And i also calculated out if the we use $e$ as the coefficient of restituition, the sum to infinity distance will be $$S_{\infty} = \dfrac{1+e^2}{1-e^2}H$$
And I also know that the total distance travelled just before the 5th bounce to be $$3.95903H$$
However i tried a few ways to get the range and failed to do so, any suggestions
Before the 5th jump, ball is at the height $e^8 H$. With the new coefficient $g$, at infinity, the total distance traveled is $\frac{1+g^2}{1-g^2}e^8H$ and it is given that total distance is larger than $4H$. i.e. \begin{equation} \frac{1+g^2}{1-g^2}e^8H > (4-3.79126)H \implies 0.329862 < g < 1\quad (g \text{ positive}) \end{equation} Since $g<e$, \begin{equation} 0.329862 < g < 0.8 \end{equation}