(i) Using the formula for power, \(P = Fv\), where \(F\) is the driving force and \(v\) is the velocity, we have:

\(144000 = F \times 16\)

\(F = \frac{144000}{16} = 9000 \text{ N}\)

Using Newton's second law, \(F - 4800 = 12500a\), where \(a\) is the acceleration at A:

\(9000 - 4800 = 12500a\)

\(4200 = 12500a\)

\(a = \frac{4200}{12500} = 0.336 \text{ m s}^{-2}\)

To find the speed at B, use the equation \(v^2 = u^2 + 2as\):

\(v^2 = 16^2 + 2 \times 0.096 \times s\)

Given \(a = 0.096 \text{ m s}^{-2}\) at B, solve for \(v\):

\(v = 24 \text{ m s}^{-1}\)

(ii) Work done by the driving force is \(5800 \times 500\).

Work done against resistance is \(4800 \times 500\).

Loss in kinetic energy is \(\frac{1}{2} \times 12500 \times (24^2 - 16^2)\).

Using energy conservation:

\(5800 \times 500 = 12500gh + \frac{1}{2} \times 12500 \times (24^2 - 16^2) + 4800 \times 500\)

Solving for \(h\):

\(5800 \times 500 = 12500gh + 4800 \times 500 + \frac{1}{2} \times 12500 \times (24^2 - 16^2)\)

\(2900000 = 12500gh + 2400000 + 12500 \times 80\)

\(2900000 = 12500gh + 2400000 + 1000000\)

\(2900000 = 12500gh + 3400000\)

\(12500gh = 2900000 - 3400000\)

\(12500gh = -500000\)

\(h = \frac{-500000}{12500 \times 9.8}\)

\(h = 20 \text{ m}\)