(i) Using the equation of motion: \(0 = u^2 - 2gs\), where \(s = 45 \text{ m}\) (40 m + 5 m) and \(g = 10 \text{ ms}^{-2}\), we have:

\(u^2 = 2 \times 10 \times 45\)

\(u = 30 \text{ ms}^{-1}\)

(ii) Using \(s = ut - \frac{1}{2}gt^2\) with \(s = 40\), \(u = 30\), and \(g = 10\):

\(40 = 30t - 5t^2\)

Solving \(5t^2 - 30t + 40 = 0\) gives \(t = 2\) and \(t = 4\).

The time above the top of the cliff is \(4 - 2 = 2 \text{ s}\).

(iii) For the second signal, the maximum height above the top of the cliff is \(\frac{1}{2}g(17 \div 4) = 21.25 \text{ m}\).

Using \(0 = V^2 - 2g(40 + 21.25)\):

\(V^2 = 2 \times 10 \times 61.25\)

\(V = 35 \text{ ms}^{-1}\)