(i) To find the speed of the ring as it reaches B, use the equation of motion:

\(v^2 = u^2 + 2as\)

where \(u = 0\) (initial speed), \(a = 2.5\) m/s² (acceleration), and \(s = 5\) m (distance).

\(v^2 = 0 + 2 \times 2.5 \times 5\)

\(v^2 = 25\)

\(v = 5\) m/s

(ii)(a) To find the speed of the ring at C, use energy conservation:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 \times \sin 30^\circ = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(6 = 0.1(v^2 - 25)\)

\(6 = 0.1v^2 - 2.5\)

\(8.5 = 0.1v^2\)

\(v^2 = 85\)

\(v = 9.22\) m/s

(ii)(b) The greatest speed of the ring occurs at the lowest point of the semicircle:

\(Potential energy loss = Kinetic energy gain\)

\(0.2 \times 10 \times 6 = 0.5 \times 0.2 \times (v^2 - 5^2)\)

\(12 = 0.1(v^2 - 25)\)

\(12 = 0.1v^2 - 2.5\)

\(14.5 = 0.1v^2\)

\(v^2 = 145\)

\(v = 12.0\) m/s