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
Create a free account to view solution
View Solution FreeMore Kinematics Questions
The velocity - time graph of a linear motion is shown in figure. The displacement & distance from the origin after 8 sec...A particle is projected from a point on the ground with an initial velocity of u = 50 m/s at an angle of 53o with the ho...A bar magnet M is allowed to fall towards a fixed conducting ring C. If g is the acceleration due to gravity, v is the v...The position of a particle moving along x-axis depends on time as x = 2(1 - e-3t). Choose the INCORRECT statement....A particle moves in east direction with 15 m/sec. for 2 sec. then moves northward with 5 m/sec. for 8 sec. then average ...