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Question

# A hot air balloon is rising vertically upwards at a constant velocity of 10 msâˆ’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 msâˆ’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 sWhen the parachutist bails out, he has the velocity of the balloon and has an upward velocity of 10 ms−1,i.e.u=+10 ms−1. Also g=−10 ms−2 (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 =45–15=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 ms−1 (directed downwards) At t=3 s, the velocity of parachutist is u2=−20 ms−1 Let v2 be the velocity with which the parachutist will reach the ground. Using third equation of motion, we know v22−u22=2aS ⇒v22−(−20)2=2(+5)(−30) ⇒v22=400−300=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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