Question

A parachutist drops freely from helicopter for 20s before the parachute opens out. Then he descends with retardation of 40 m/s^2. The height of the helicopter from the ground level is 2500m. When he drops out his velocity on teaching the ground will be

Answer 1

The initial velocity (u) of the parachutist is 0

Now V=u+gt

=> V=0+(-10)10

=> V = -100

Then, S=ut+1/2ut^2

=> S=0+1/2(-10)*100

=> S = 500

=> H=2495-500

=> H =1995

We know that,

 v^2=u^2+2ah=> v^2=10000-2*2.5*1995=> v^2 = 25 => v = 5 m/s
Hence,when he drops out his velocity on teaching the ground will be5 m/s.