Explanation

(c) Parachute bails out at height $H$ from ground. Velocity at $A$
$v = \sqrt{2\ gh}$
$\ \ \ = \sqrt{2 \times 9.8 \times 50} = \sqrt{980}{ms}^{- 1}$
The velocity at ground $v_{1} = 3\ {ms}^{- 1}$ (given)
$Acceleration = - 2\ {ms}^{- 2}$ (given_
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$\therefore H - h = \frac{v^{2} - v_{1}^{2}}{2 \times 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{980 - 9}{4} = \frac{971}{4} = 242.75$
$\therefore\ \ \ \ \ \ \ H = 242.75 + h$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 242.75 + 50 = 293\ m$