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, \(a = 9.8 \; m/s^{2}\) , 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

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

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

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

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