a
(a) After bailing out from point \(A\) parachutist falls freely under gravity. The velocity acquired by it will \(‘v’\)

From \({v^2} = {u^2} + 2as\) \( = 0 + 2 \times 9.8 \times 50\) \(= 980\)

[As \(u = 0\), \(a = 9.8m/{s^2}\), \(s = 50\, m\)]

At point \(B\), parachute opens and it moves with retardation of 2\(m/{s^2}\) and 

reach at ground (Point \(C\)) with velocity of \(3\,m/s\)

For the part \(‘BC’\) by applying the equation \({v^2} = {u^2} + 2as\)

\(v = 3\,m/s\), \(u = \sqrt {980} \,m/s\), \(a = - 2m/{s^2}\), \(s = h\)

\(⇒  {(3)^2} = {(\sqrt {980} )^2} + 2 \times ( - 2)\, \times \,h ⇒  9 = 980 - 4h\)

\(⇒  h = \frac{{980 - 9}}{4}\) \( = \frac{{971}}{4} = 242.7 \tilde = 243\,m\).

So, the total height by which parachutist bail out = \(50 + 243 = 293\, m\).