(i) Apply Newton's second law to particle A:

\(4 - T = 0.4a\)

Apply Newton's second law to particle B:

\(T - 2 = 0.2a\)

For the system:

\(4 - 2 = (0.4 + 0.2)a\)

Solving the system of equations gives:

\(a = \frac{10}{3} \text{ m s}^{-2}, \; T = \frac{8}{3} \text{ N}\)

(ii) Use \(v^2 = u^2 + 2as\) for particle B with \(a = \frac{10}{3}\):

\(v^2 = 0 + 2 \times \frac{10}{3} \times 0.5\)

\(v^2 = \frac{10}{3}\)

After A hits the ground, apply \(v^2 = u^2 + 2as\) to find \(s\):

\(0 = \frac{10}{3} - 2 \times 10 \times s\)

\(s = \frac{1}{6} \text{ m}\)

Maximum height = \(\frac{1}{2} + \frac{1}{2} + \frac{1}{6} = \frac{7}{6} = 1.17 \text{ m}\)