Explanation

(a) After balling out from point $A$ parachutist falls freely under gravity. The velocity acquired by it will ‘$v$’
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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 $v^{2} = u^{2} + 2as$
$v = 3m/s,\ u = \sqrt{980}\ m/s,\ a = - 2m/s^{2},\ s = h$
$\Rightarrow (3)^{2} = \left( \sqrt{980} \right)^{2} + 2 \times ( - 2) \times h \Rightarrow 9 = 980 - 4h$
$\Rightarrow h = \frac{980 - 9}{4} = \frac{971}{4} = 242.7 \cong 243\ m$
So, the total height by which parachutist bail out $= 50 + 243 = 293\ m$