InterviewSolution
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- An astronaut jumps from an aeroplane. After he hadfallen 40 m, then his parachute opens. Now he falswith a retardation of 2.0 m/s and reaches the eariwith a velocity of 3.0 m/s. What was the height of theaeroplane? For how long astronaut remained in air |
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Answer» height of aeroplane is approx.=233mhe remains in the air for approximately =15s plz give solution is it correct? yes answer is correct ..but how to solve hey saurabh can u plz tell that is my answer correct i will provide you solution for it wait i am sending sol. yes plzz..send hope u understand Let the inital distance traveleld before opening parachute be h1 = 40m then s = ut + (1/2)gt2 as u =0 40= (1/2)g.(t1)2 or t1 = √80/9.8 = √400/49 = 20/7 s his velocity at that time is v1 = u + gt1 = 10 X 20/7 = 28.28 m/s Then there is retardation of 2 m/s v22- v12= 2gh2 so 32= (28.28)2- 2g.h2 or h2 = 25x31/(2x9.8) Total height will be h = h1 + h2 = h2+40 For calculating the time after parachute opens, v2 = v1 - at 3 = 28 - 2t2, or t2=25/2 Total time, t = t1 + t2 = 25/2+20/7 ortotal time of flight would be t = 15.35 s |
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