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

\(mgh = 0.5 \times 10 \times 1.8\)

The kinetic energy at B is:

\(\frac{1}{2} mv^2\)

Equating the potential energy to kinetic energy:

\(\frac{1}{2} v^2 = 10 \times 1.8\)

\(v^2 = 36\)

\(v = 6 \text{ m s}^{-1}\)

(ii) The work done against resistance is the loss of kinetic energy as the particle moves from B to C. The kinetic energy at B is:

\(\frac{1}{2} \times 0.5 \times 6^2\)

The kinetic energy at C is:

\(\frac{1}{2} \times 0.5 \times 5^2\)

The work done is the difference:

\(\text{WD} = \frac{1}{2} \times 0.5 \times (6^2 - 5^2)\)

\(\text{WD} = 2.75 \text{ J}\)