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

\(0.9g - T = 0.9a\)

For particle B:

\(T - 0.6g = 0.6a\)

Adding the two equations:

\(0.9g - 0.6g = 0.9a + 0.6a\)

\(0.3g = 1.5a\)

Solving for \(a\):

\(a = \frac{0.3g}{1.5} = 2 \text{ m/s}^2\)

Substituting \(a\) back to find \(T\):

\(T = 0.6g + 0.6 \times 2 = 0.6 \times 9.8 + 1.2 = 7.2 \text{ N}\)

(ii) After A hits the floor, B moves upwards for 0.3 s with initial velocity \(u = 3 \text{ m/s}\) (from previous motion):

Using \(v = u - gt\):

\(0 = 3 - 9.8 \times 0.3\)

Solving for \(t\):

\(t = 0.3 \text{ s}\)

Using \(v^2 = u^2 + 2ah\) to find height \(h\):

\(0 = 3^2 - 2 \times 9.8 \times h\)

\(9 = 19.6h\)

\(h = \frac{9}{19.6} = 0.459 \text{ m}\)

Total height from initial position:

\(h = 2.25 \text{ m}\)