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Question

A hot air balloon is rising vertically upwards at a constant velocity of 10 ms−1. When it is at a height of 45 m from the ground, a man bails out from it. After 3 s he opens his parachute and decelerates at a constant rate of 5 ms−2. After how long does the parachutist takes to hit the ground after his exit from the balloon?

A
4 s
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B
5 s
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C
6 s
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D
7 s
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Solution

The correct option is B 5 s
When the parachutist bails out, he has the velocity of the balloon and has an upward velocity of 10 ms1,i.e.u=+10 ms1.
Also g=10 ms2 (acting downwards).
The displacement of parachutist in t=3 s is given by
S=ut+12gt2 (By second equation of motion)
=10×3+12×(10)×(3)2
=15 m
Since the displacement is negative, it is directed downwards.
So the height of the parachutist from the ground when he opened his parachute =4515=30 m.
The velocity of the parachutist 3 s after he bails out is
u2=u+gt (Using first equation of motion)
=10+(10)×3=20 ms1 (directed downwards)
At t=3 s, the velocity of parachutist is u2=20 ms1
Let v2 be the velocity with which the parachutist will reach the ground.
Using third equation of motion, we know
v22u22=2aS
v22(20)2=2(+5)(30)
v22=400300=100
v2=10 m/s (directed downwards)
Let t be the time taken by the parachutist to reach ground after he opens up his parachute Using first equation of motion, we can say
v2=u2+at
10=20+(5)t
t=2 s
The total time the parachutist takes (after his exit from the balloon) to hit the ground is =3 s+2 s=5 s
Thus, the correct choice is (b).

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