(a) \(T-30=3a,\quad50-T=5a\). Thus \(a=20/8=2.5\,\mathrm{m\,s^{-2}}\) and \(T=37.5\) N. B falls 2 m: \(2=\frac12(2.5)t^2\), so \(t=\sqrt{1.6}=1.26\) s.

(b) When B hits the plane, A is at height 3 m and has upward speed \(u=\sqrt{2(2.5)(2)}=\sqrt{10}\). The string becomes slack. At time \(\tau\) after this instant, \[h=3+\sqrt{10}\tau-5\tau^2.\] Setting h=3.25 gives \[5\tau^2-\sqrt{10}\tau+0.25=0,\quad \tau=\frac{\sqrt{10}\pm\sqrt5}{10}.\] Between the two roots A is at least 3.25 m high. Duration \(=2\sqrt5/10=\sqrt5/5=0.447\) s.