(i) The kinetic energy (KE) at A is equal to the loss of potential energy (PE) from O to A. The vertical height from O to A is given by:

\(h = 2.4 \sin 50^\circ\)

The loss of PE is:

\(\text{PE loss} = mgh = 0.8 \times 9.8 \times 2.4 \sin 50^\circ\)

\(\text{KE at A} = \text{PE loss} = 14.7 \text{ J}\)

(ii) Using the kinetic energy at C, which is equal to the kinetic energy at A:

\(\frac{1}{2} mv^2 = 14.7\)

\(v^2 = \frac{2 \times 14.7}{0.8}\)

\(v = \sqrt{36.75} \approx 6.06 \text{ m s}^{-1}\)

(iii) Using the principle of conservation of energy, the greatest speed is 8 m s^-1:

\(\frac{1}{2} m (8^2) = mgh + \frac{1}{2} m (6.06^2)\)

\(32 = gh + 18.36\)

\(gh = 32 - 18.36 = 13.64\)

\(h = \frac{13.64}{9.8} \approx 1.36 \text{ m}\)