(i) Using conservation of energy, the potential energy lost is converted into kinetic energy. The potential energy at A is given by:

\(mgh = 40 \times g \times 7.2\)

The kinetic energy at B is:

\(\frac{1}{2} \times 40 \times v^2\)

Equating the potential energy and kinetic energy:

\(\frac{1}{2} \times 40 \times v^2 = 40 \times g \times 7.2\)

Solving for \(v\):

\(v = 12 \text{ m s}^{-1}\)

(ii) For the rough slide, the work done against friction (WDF) is the same in both cases. First, calculate WDF for the first scenario:

\(\text{WDF} = 40 \times g \times 7.2 - \frac{1}{2} \times 40 \times 10^2 = 880\)

For the second scenario, using the work-energy principle:

\(\frac{1}{2} \times 40 \times V^2 + 40 \times g \times 7.2 = \frac{1}{2} \times 40 \times 11^2 + 880\)

Solving for \(V\):

\(V = \sqrt{21} = 4.58 \text{ m s}^{-1}\)