KinematicsHard
Question
A balloon is rising upwards at a velocity of 10m/s. When it is at a height of 45 m from the ground, a parachute bails out from it. After 3s the man with the parachute decelerates at a constant rate of 5 m/s2. What was the height of the man above the ground when he opens his parachute ? (Take g = 10 m/s2).
Options
A.60 m
B.45 m
C.30 m
D.15 m
Solution
When the man exits from the balloon, his velocity is 10 m/s upwards i.e., u = + 10 m/s,
After 3s, the man reaches to a height
h = ut +
gt2
= 10 × 3 -
× 10 × 32
(Here, g = - 10 m/s2)
or h = - 15 m
Negative sign indicates that the displacement is directed downwards.
Therefore, the height of the man above the ground, when he opens his parachute i.e., after 3s, is
= 45 - 15 = 30 m
After 3s, the man reaches to a height
h = ut +
= 10 × 3 -
(Here, g = - 10 m/s2)
or h = - 15 m
Negative sign indicates that the displacement is directed downwards.
Therefore, the height of the man above the ground, when he opens his parachute i.e., after 3s, is
= 45 - 15 = 30 m
Create a free account to view solution
View Solution FreeMore Kinematics Questions
A body starting from rest moves along straight line with a constant acceleration. The variation of speed (v) with distan...There are two values of time for which a projectile is at the same height. The sum of these two times is equal to (T = t...The time dependence of a physical quantity P is given by P = P0 exp (−α t2), where a is a constant and t is t...The relation between time t and distance x is t = ax2 + bx where a and b are constants.The acceleration is...A, B, C and are points in a vertical line such that AB = AC = CD. If a body falls from rest A, then the times of descend...