Question

A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he de-
scends with a net retardation of 2.5 m/s². If he bails out of the plane at a height of 2495 m and g = 10
m/s-, hit velocity on reaching the ground will be:
(A) 5 m/s
(B) 10 m/s
(C) 15 m/s
(D) 20 m/s
​

Answer 1

Answer:

Explanation: the Velocity will be 20m/s

Answer 2

Answer:

(A)5m/s

Explanation:

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