(a) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 24 \text{ m}\), \(t = 2 \text{ s}\), and \(a = -g = -9.8 \text{ m s}^{-2}\).

\(24 = u \times 2 - \frac{1}{2} \times 9.8 \times 2^2\)

\(24 = 2u - 19.6\)

\(2u = 43.6\)

\(u = 22 \text{ m s}^{-1}\)

(b) At maximum height, \(v = 0\), using \(v^2 = u^2 + 2as\):

\(0 = 22^2 - 2 \times 9.8 \times s\)

\(s = 24.2 \text{ m}\)

Height down in 1.8 s: \(s = \frac{1}{2} \times 9.8 \times 1.8^2 = 16.2 \text{ m}\)

\(h = 24.2 - 16.2 = 8 \text{ m}\)

Alternative method: Using \(v = u - gt\) with \(u = 22\) and \(v = 0\):

\(0 = 22 - 9.8t\)

\(t = 2.2 \text{ s}\)

\(h = 22 \times (2.2 - 1.8) - \frac{1}{2} \times 9.8 \times (2.2 - 1.8)^2\)

\(h = 8 \text{ m}\)