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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

Answer»

height of aeroplane is approx.=233mhe remains in the air for approximately =15s

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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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