Question

A parachutist falls freely from the plane for first 10 seconds.Then it decelerates at a rate of 2.5m/s^2 .If he bails out of the plain from a height of 2495m then what is the velocity on reaching the ground.a=10

Answer 1

The initial velocity of the parachutist is zero

now V=u+gt

V=0+(-10)10

V=-100 (-sign as the velocity is in downward direction)

Now S=ut+1/2ut^2

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

S=500

Now H=2495-500

=1995

NOw v^2=u^2+2ah

v^2=10000-2*2.5*1995

v^2=25

v=5

so his velocity on reaching the ground is 5m/s