Apply Newton's second law to the system:

For particle A: \(2g - T = 2a\)

For particle B: \(T - 3g \sin 18 = 3a\)

Solving these equations gives:

\(2g - 3g \sin 18 = 5a\)

\(a = 2.145898034\)

Use the equation \(v^2 = 2as\) to find \(v^2\) when A reaches the ground:

\(v^2 = 2 \times 2.145898034 \times 0.45 = 1.931308\)

\(v = 1.389715162\)

When the string becomes slack, find the acceleration \(a\) for the motion of B:

\(T = 0, 3g \sin 18 = 3a\)

\(a = 3.0901699\)

Use \(v^2 = 2as\) to find the further distance \(s\) travelled by B:

\(0 = 1.93 - 2 \times 3.09 \times s\)

\(s = 0.312\)

Total distance moved by B is \(0.45 + 0.312 = 0.762\) m.