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A 60.0-kg and a 90.0-kg skydiver jump from an airplane at an altitude of 6.00×103m , both falling in a headfirst position. Make some assumption on their frontal areas and calculate their terminal velocities. How long will it take for each skydiver to reach the ground (assuming the time to reach terminal velocity is small)? Assume all values are accurate to three significant digits.

Respuesta :

Answer:

  • The terminal velocity for each sky diver is 342.92 m/s
  • The time taken is 34.99 s

Solution:

As per the question:

Mass of the skydiver 1 = 60.0 kg

Mass of the skydiver 2 = 60.0 kg

Altitude, H = [tex]6.00\times 10^{3}\ m[/tex]

Now,

To calculate the terminal velocities of the skydivers:

We know that the drag force equals the weight of the diver and the net force acting on the body of the divers is zero as the body has constant velocity.

This velocity is the terminal velocity of the body.

Now,

Using the law of conservation of energy:

Potential Energy, PE = Kinetic Energy, KE

[tex]mgH = \frac{1}{2}mv_{t}^{2}[/tex]

Thus

[tex]v_{t} = \sqrt{2gH} = \sqrt{2\times 9.8\times 6.00\times 10^{3}} = 342.92\ m/s[/tex]

Now,

The time taken by each skydiver to reach the ground is given by:

[tex]H = ut + \frac{1}{2}gt^{2}[/tex]

where

u = initial velocity = 0 m/s

Thus

[tex]t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2\times 6.00\times 10^{3}}{9.8}}[/tex]

t = 34.99 s

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