A parachutist after bailing out falls 50 m without friction when parachute opens, it decelerates at 2 ms-2. He reaches the ground with speed of 3 m/ s. At what height did he bail out?

After bailing out from point A, parachutist falls freely under gravity. The velocity acquired by it will be v.From v2 = u2 + 2as = 0 + 2 x 9.8 x 50 = 980                                        (As, u = 0, a = 9.8 ms-2 , s = 50 m)At point B, parachute opens and it moves with retardation of 2 ms-2 and reach at ground (point C with velocity of 3 ms-1 ).For the part BC by applying the equation v2 = u2 + 2as                v=3 ms-1,u=980 ms-1,a=-2 ms-2,s=h⇒      (3)2=(980)2+2×(-2)×h⇒         9=980-4h⇒        h=980-94=9714=242.7≅243 mSo, the total height by which parachutist bails out              = 50 + 243 = 293 m