(a) The sequence is a geometric progression with first term \(a = 50\) and common ratio \(r = \frac{40}{50} = 0.8\).

The general term is given by \(a_n = 50 \times 0.8^{n-1}\).

We need to find the smallest \(n\) such that \(a_n < 10\).

\(50 \times 0.8^{n-1} < 10\)

\(0.8^{n-1} < 0.2\)

Using logarithms, \((n-1) \log(0.8) < \log(0.2)\)

\(n-1 > \frac{\log(0.2)}{\log(0.8)} \approx 7.2\)

\(n > 8.2\), so \(n = 9\).

Hence, the 9th impact is required.

(b) The sum of the first 20 terms of a geometric progression is given by:

\(S_{20} = \frac{50(1 - 0.8^{20})}{1 - 0.8}\)

\(S_{20} = \frac{50(1 - 0.8^{20})}{0.2}\)

\(S_{20} = 50 \times 5 \times (1 - 0.8^{20})\)

\(S_{20} \approx 247\) mm (to the nearest millimetre).

(c) The greatest total depth is the sum to infinity of the geometric progression:

\(S_\infty = \frac{50}{1 - 0.8}\)

\(S_\infty = \frac{50}{0.2}\)

\(S_\infty = 250\) mm.