એક પેરાશૂટધારી કૂદી પડયા પછી ઘર્ષણરહિત અવસ

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 = $50 + 243 = 293\, m$.