(i) To find the maximum height of A, use the equation for vertical motion:

\(v = u - gt\)

where \(v = 0\) at maximum height, \(u = 5 \text{ m/s}\), and \(g = 9.8 \text{ m/s}^2\). Solving \(0 = 5 - 9.8t\) gives \(t = \frac{5}{9.8}\).

The height \(h\) is given by:

\(h = ut - \frac{1}{2}gt^2\)

Substitute \(t = \frac{5}{9.8}\) to find \(h_A\).

For B, use the same formula with \(u = 8 \text{ m/s}\) and the same \(t\). Calculate \(h_B\).

The difference in heights is \(h_B - h_A = 1.5 \text{ m}\).

(ii) To find when B is 0.9 m above A, solve:

\(h_B - h_A = 0.9\)

Using \(h = ut - \frac{1}{2}gt^2\) for both particles, set up the equation:

\(8t - \frac{1}{2}gt^2 - (5t - \frac{1}{2}gt^2) = 0.9\)

Simplify to \(3t = 0.9\), giving \(t = 0.3 \text{ s}\).

Substitute \(t = 0.3\) into \(h_A = 5t - \frac{1}{2}gt^2\) to find \(h_A = 1.05 \text{ m}\).