(i) Using the equation of motion: \(s = ut + \frac{1}{2} a t^2\), where \(s = 0.36\) m, \(u = 0\), and \(t = 0.6\) s. Substituting, we get:

\(0.36 = 0 + \frac{1}{2} a (0.6)^2\)

\(0.36 = 0.18a\)

\(a = \frac{0.36}{0.18} = 2 \text{ m/s}^2\)

(ii) Applying Newton's second law to particle A: \(0.45g - T = 0.45a\)

\(0.45 \times 10 - T = 0.45 \times 2\)

\(4.5 - T = 0.9\)

\(T = 4.5 - 0.9 = 3.6 \text{ N}\)

(iii) Applying Newton's second law to particle B: \(T - mg = ma\)

\(3.6 - m \times 10 = m \times 2\)

\(3.6 = 12m\)

\(m = \frac{3.6}{12} = 0.3 \text{ kg}\)

(iv) Using the equation \(v^2 = u^2 + 2as\) for particle B when it reaches maximum height (\(v = 0\)):

\(0 = (0.6 \times 2)^2 - 2 \times 10 \times s\)

\(0 = 1.44 - 20s\)

\(s = \frac{1.44}{20} = 0.072 \text{ m}\)

Maximum height above the floor = initial height + \(s = 0.72 + 0.072 = 0.792 \text{ m}\)