(i) The power required to maintain a constant speed is given by the formula:

\(P = Fv\)

where \(F = 3000 \text{ N}\) and \(v = 30 \text{ m s}^{-1}\).

Thus, \(P = 3000 \times 30 = 90,000 \text{ W} = 90 \text{ kW}\).

(ii) Using the energy method, the potential energy gained is:

\(PE = 25000gh\)

The initial kinetic energy at A is:

\(KE_{\text{initial}} = \frac{1}{2} \times 25000 \times 30^2 = 11,250,000 \text{ J}\)

The final kinetic energy at B is:

\(KE_{\text{final}} = \frac{1}{2} \times 25000 \times 25^2 = 7,812,500 \text{ J}\)

The loss in kinetic energy is:

\(KE_{\text{loss}} = 11,250,000 - 7,812,500 = 3,437,500 \text{ J}\)

Using the work-energy principle:

\(KE_{\text{initial}} = KE_{\text{final}} + 25000gh + \frac{3000h}{\sin 2^\circ}\)

Solving for \(h\):

\(3,437,500 = 25000gh + \frac{3000h}{\sin 2^\circ}\)

\(h = 10.2 \text{ m}\)