Let the acceleration of the system be \(a\). Using Newton's second law for both particles:

For particle A: \(T - 0.2g = 0.2a\)

For particle B: \(0.6g - T = 0.6a\)

Adding these equations gives:

\(0.6g - 0.2g = 0.6a + 0.2a\)

\(0.4g = 0.8a\)

\(a = \frac{0.4g}{0.8} = 5 \, \text{m/s}^2\)

When B reaches the floor, using \(v^2 = u^2 + 2as\) with \(u = 0\), \(a = 5\), and \(s = 1.6\):

\(v^2 = 2 \times 5 \times 1.6\)

\(v^2 = 16\)

\(v = 4 \, \text{m/s}\)

For A to reach 3 m above the floor, the total distance moved by A is \(3 - h\). Using energy conservation:

\(Potential energy gain = Kinetic energy loss\)

\(0.2g(3 - h) = \frac{1}{2} \times 0.2 \times 4^2\)

\(0.2g(3 - h) = 1.6\)

\(2g(3 - h) = 16\)

\(3 - h = 0.8\)

\(h = 3 - 0.8 = 0.6\)