Geometric series, distance travelled when the ball hits the ground for the 5th time

Question

a ball dropped on the surface takes a sequence of vertical bounces with each bounce the ball loses 15ft its preceding height. the ball is dropped from 25 feet. find the total distance travelled when the ball hits the ground for the 5th time

Answer 1

The question can be formulated with respect to heights travelled, and with respect to an elasticity constant $e \in (0,1)$ which tells us how much energy has been lost due to the bounce.

We will denote by $h_n$ the maximum heights of the ball after the $n$th bounce, where we will assume that $h_0 = 25$ and we know that $h_1 = 10$ (after having lost 15ft after its first bounce).

The only way we can proceed now is by knowing/approximating how much energy is lost after each bounce, and we'll quantify this using an elasticity constant $e_n$ for the proportion of energy lost of the $n$th bounce. We can write this as $e_n = \frac{h_{n+1}}{h_{n}}$.

We will assume that the proportion of energy lost after each consecutive bounce is equal for all bounces, and, and thus can assume that $e_n$ is equal throughout the whole system (so we will drop the subscript $n$ for brevity).

Knowing the values of $h_1$ and $h_0$, we can see that $e = \frac{h_1}{h_0} = \frac{10}{25} = 0.4$, therefore we may write $$h_{n+1} = e h_n = e^{n+1}h_0.$$

To calculate the total distance travelled when the ball hits the ground for the kth time, we need to now count the total number of times the max height is traversed. Indeed, setting $D_k$ to be the parameter for the distance travelled by the ball when hitting the ground for the $k$th time, we have that for $k=1$ $$ D_1 = h_0 $$ and for $k\geq 2$ $$ D_k = h_0 + 2\sum_{i=1}^{k-1}h_i $$ Substituting the formula for $h_n$, we find that $$ \begin{align} D_k &:= h_0 + 2\sum_{i=1}^{k-1}h_i \\ &= h_0 + 2h_0\sum_{i=1}^{k-1}e^i \\ &= h_0 + 2h_0 \frac{e-e^k}{1-e} \\ &= h_0\left(1 + \frac{2(e-e^k)}{1-e}\right) \\ &= h_0\left(\frac{1 + e - 2e^{k}}{1-e}\right) \end{align} $$ Plugging in $k=5$ then yields $$ D_5 = 57.48 \text{ft} $$ as being the total distance travelled by the ball after hitting the floor for the 5th time.