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Home / Doubts and Solutions/JEE/Physics

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Asked by pukai7229 16th May 2018, 9:30 AM
Answered by Expert
Answer:
When the parachutist bails out from the balloon, his initial speed is 10 m/s upwards (due to balloon vertical speed )
 
Hence the distance h travelled in 3 seconds before he opens parachute is calculated as shown below
 
h = (-10)×3 +(1/2)×10×9 = 15 m.
 
when the parachutist bails out from the balloon, the height was 45 m above ground. Hence at the time of openeing the parachute, parachutist will be at (45-15) = 30 m above the ground. (question(a) answered).
 
During this 3 seconds the balloon has ascended to a distance 3×10 = 30 m. Hence the parachutist is at (30+15)=45 m from the balloon when he opens parachute ( question (b) answered)
 
When the parachutist opens his parachute, his speed at that instant is given by
 
v = -10+10×3 = 20 m/s
 
To strike the ground he has to travel 30 m vertically downwards. His speed vg when he reaches ground level is given by
 
vg2 = 20×20-(2×5×30) = 100; hence vg = 10 m/s (question (c) answered)
 
Time taken is calculated as follows
 
vg = v-at ;   10 = 20-(5×t) 
 
solving the above equation, we get t=2 seconds. This time is from the instance of opening parachute to the instance of reaching the ground. Before opening parachute, he has spent 3 seconds from the instance of bailing out from the balloon. Hence total time for the parachutist to reach ground is 5 seconds
Answered by Expert 16th May 2018, 12:12 PM
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