A parachutist after bailing  out fall for 5 sec and when it open he descends with an deceleration of 1 m/s2 and reach the ground with a speed of 3 m/s .Find the height at which he was dropped.

till up to 5 sec  he is like a freely falling bodywe know, S=Ut+12at2, here displacement=S;initial velocity=U; time=t; acceleration=a; up to 5 sec it is like a freely falling body     S=0+12g52 S=12×10×5×5=125m The velocity parachutist  gain  at t=5sec V=U+at   ⇒V=0+10×5 ⇒V=50m/sAfter the parachute is open initial velocity U=50m/s;final velocity   V=3m/s; acceleration     a=−1m/s2 ; displacement=  S=?  we know,V2−U2=2as substitute the values in above equation32−502=2−1S ⇒S=1245.5m Total height is 1245.5+125=1370.5m