A parachutist after bailing out falls 50m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height from ground did he bail out?

given initial velocity of parachutist=u=0; distance=s=50m;acceleration due to gravity=g=9.8m/s2substitute given values in,       v2−u2=2as     v2=2×9.8×50    v2=980 m/s---(1)v=The velocity after he descends 50m after 50m parachute opens;  final velocity =v=3m/s; a=-2m/s2 initial velocity =u⇒u2=980 substitute, to obtain distance,  in v2−u2=2as 32-980=2(-2)s s=9724=243m The height at which he bailed out =h=243+50=293m