Question

An astronaut jumps from an aeroplane after he had fallen 40 metre then his parachute opens now he falls with a retardation of 2.0 metre per second square and reaches the earth with a velocity of 3.0 metre per second what was the height of the aeroplane for how long astronaut remained in air

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

Answer 2

Answer:

The height of the aeroplane is 233.75m and total time remained in air is 15.36s.

Explanation:

Given an astronaut jumps from an aeroplane, and fallen 40 m before opening parachute, $h=40m$

Since it is a case of freely falling body, initial velocity, $u=0$

Using the equation of motion, for acceleration due to gravity, $g=9.8m/s^{2}$

$v^{2} - u^{2} =2gh$

$\implies v^{2} =2\times 9.8\times40\implies v^{2} =784$

$\implies v=\sqrt{784} =28 m/s$

So, velocity through 40m is 28m/s.

Then time taken is, from the equation of motion

$v=u+gt_1$

$\implies 28=9.8t_1\implies t_1=\dfrac{28}{9.8} \approx2.86s$

After opening the parachute, given that there is a retardation, $a=- 2 m/s^{2}$

Reaches the earth with velocity, $v_2=3 m/s$

The time taken after opening the parachute is

$v_2=v+at_2$

$\implies3=28+(-2)t_2\implies3-28=-2t_2$

$\implies -25=-2t_2\implies t_2=\dfrac{25}{2} =12.5s$

Using the equation of motion in this case

$v_2^{2} - v^{2} =2ah_1$

$\implies 3^{2} - 28^{2} =2\times(-2)h_1$

$\implies 9 - 784=-4h_1$

$\implies h_1=\dfrac{ - 775}{-4} =193.75m$

Height of the aeroplane is the sum of the distance travelled before opening the parachute and after opening the parachute

$H=h+h_1=40+193.75=233.75m$

Total time in air is the sum of the time before opening the parachute and after opening the parachute

$t = t_1 + t_2 = 2.86+12.5=15.36s$

Thus, the height of the aeroplane is 233.75m and total time remained in air is 15.36s.

For more problems

https://brainly.in/question/13174366

https://brainly.in/question/18038405