I'm having a bit of trouble determining what I assume to be an infinite geometric sum for the following question in part ii.)
I've posted my working below, but I'm not sure how to get the geometric sum in the form the question requires.
Working for Geometric Sum Problem:
You're forgettingthat the ball first falls with height H, then rises to height $\frac{qH}{100}$, falls with height $\frac{qH}{100}$ and so on.
Therefore, the correct distance must be :
H + 2(S)
where S is the sum of heights starting from $\frac{qH}{100}$.
$S = \frac{\frac{qH}{100}}{1 - \frac{q}{100}}$
Simplifying, $S = \frac{qH}{100 - q}$
Going back to the original problem,
$H + 2(\frac{qH}{100 - q})$
$\frac{100H - qH + 2qH}{100 - q}$
$\frac{100H + qH}{100 - q}$
$H(\frac{100 + q}{100 - q})$
which is the required answer.