(i) Using the equation of motion:

\(s = \frac{1}{2} (u + v) t\)

where \(s = 200\) m, \(u = 0\), \(t = 10\) s, we have:

\(200 = \frac{1}{2} (0 + v) \times 10\)

\(v = 40 \text{ m s}^{-1}\)

To find acceleration \(a\), use:

\(s = ut + \frac{1}{2} a t^2\)

\(200 = 0 \times 10 + \frac{1}{2} a \times 10^2\)

\(a = 4 \text{ m s}^{-2}\)

(ii) After 10 s, the rocket moves under gravity alone. Using:

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

where \(v = 0\), \(u = 40 \text{ m s}^{-1}\), \(a = -g\), we have:

\(0 = 40^2 - 2 \times 9.8 \times s\)

\(s = 80 \text{ m}\)

\(Total height = 200 + 80 = 280 m\)

(iii) Time to reach maximum height:

\(0 = 40 - gt_1\)

\(t_1 = 4 \text{ s}\)

Time to fall from maximum height:

\(280 = \frac{1}{2} g t_2^2\)

\(t_2 = \sqrt{\frac{560}{9.8}} = 7.48 \text{ s}\)

\(Total time = 10 + 4 + 7.48 = 21.5 s\)