(i) Using conservation of energy, the potential energy at A is converted to kinetic energy at E. The equation is:

\(mgh = \frac{1}{2} m (6)^2\)

Solving for \(h\):

\(gh = \frac{1}{2} (6)^2\)

\(h = \frac{18}{g}\)

Assuming \(g = 9.8 \text{ m/s}^2\),

\(h = \frac{18}{9.8} \approx 1.8 \text{ m}\)

(ii) The maximum speed occurs at the lowest point C. Using energy conservation:

\(\frac{1}{2} mv^2 = mg(1.8 + 0.65)\)

\(v^2 = 2g(2.45)\)

\(v = \sqrt{2 \times 9.8 \times 2.45} \approx 7 \text{ m/s}\)