(i) To find the greatest height, use the equation of motion: \(v^2 = u^2 + 2as\). Here, \(v = 0\) at the greatest height, \(u = 6\) m/s, \(a = -g = -9.8\) m/s², and \(s = 0.5\) m.

First, find the initial velocity \(u\) when the particle is at 0.5 m:

\(6^2 = u^2 - 2 \times 9.8 \times 0.5\)

\(36 = u^2 - 9.8\)

\(u^2 = 46\)

Now, find the total height:

\(0 = 46 - 2 \times 9.8 \times s\)

\(2 \times 9.8 \times s = 46\)

\(s = \frac{46}{19.6} = 2.3\) m

(ii) To find the time to return to O, use \(s = ut + 0.5gt^2\) with \(s = 2.3\) m, \(u = 0\), and \(g = 9.8\) m/s²:

\(2.3 = 0 + 0.5 \times 9.8 \times t^2\)

\(t^2 = \frac{2.3}{4.9}\)

\(t = \sqrt{\frac{2.3}{4.9}} = 0.678\) s

Total time for the round trip is \(2 \times 0.678 = 1.36\) s