KinematicsHard
Question
A parachutist drops freely from an aeroplane for 10s before the parachute opens out. then he descends with a net retardation of 2.5 m/s2. If he bails out of plane at at of 2495 m and g = 10 m/s2, hit velocity on reaching the ground will be :-
Options
A.5 m/s
B.10 m/s
C.15 m/s
D.20 m/s
Solution
Velocity after 10 sec is equal to
v = u+ at
v = u+ at
v = 0 + (9.8) (10) = 100 m/s
Distance covered in 10 sec is equal to
s=
(10) (10)2 = 500 m
Now from v2 = u2 + 2as.
⇒ v2 = (100)2 - 2 (2.5) (2495 - 500) = 25 ⇒ v = 5 ms-1
Distance covered in 10 sec is equal to
s=
(10) (10)2 = 500 mNow from v2 = u2 + 2as.
⇒ v2 = (100)2 - 2 (2.5) (2495 - 500) = 25 ⇒ v = 5 ms-1
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