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

From \(v^{2} \quad u^{-i} + 2\alpha 5\) \(=0 + 2 \times 9.8 / 50\) = 980

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

At point B , parachute opens and it moves with retardation of 2 \(\mathfrak{m}!3^{\dot{\gamma}}\) and reach at ground (Point C ) with velocity of 3 m/s

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

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

⇒ ⇒ \((3)^2 = (\sqrt{987})^2 + 2 \times (-2) \times h\) ⇒ ⇒ 9 = 980 - 4h

⇒ ⇒ \(h = \frac{980 - 9}{4}\) \(-\frac{971}{4} - 242.7^{2} = 243\) m.

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