Question

a parachutist dropped freely from helicopter for 20 seconds before the parachute opens out . then he descends with retardation of 40 m per second square . the height of the helicopter from the ground level is 2500m. when he drops out, his velocity on reaching the ground will be
a)10
b)20
c)
$10 \sqrt{10}$
d)zero

Answer 1

The correct option is: (d) v= 0m/s.

The initial velocity of the parachutist is zero

now V=u+gt

V=0+(-10)20

V=-200 (-sign as the velocity is in downward direction)

Now S=ut+1/2ut^2

so S=0+1/2(-10)*400

S=2000

Now H=2500-2000

=500

Now v^2=u^2+2ah

v^2=40000-(2*40*500)

v^2=0

v=0 m/s.

Answer 2

Answer:

the answer is zero m/s

Explanation:

plz see into the pic..

hope it helps..

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