Question

an astronaut jumps from a plane after he had fallen 40 m then his parachute opens now he falls with retardation of 2m/s² reaches the earth with a velocity of 3m/s what's the height of the airplane? How long Astronaut remain in air? g= 9.8m/s^2a

Answer 1

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

or total time of flight would be t = 15.35 s