The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of a 79.5 kg skydiver falling in a pike (headfirst) position with a surface area of 0.170 m2. (Assume that the density of air is 1.21 kg/m3 and the drag coefficient of a skydiver in a pike position is 0.7.)

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cryssatemp cryssatemp · Answer 1 · 2019-10-08T12:15:15+02:00

Answer: 104.026 m/s=374.49 km/h

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called "drag force" [tex]D[/tex]:

[tex]D={C}_{d}\frac{\rho V^{2} }{2}A[/tex] (1)

Where:

[tex]C_ {d}=0.7[/tex] is the drag coefficient

[tex]\rho=1.21 kg/m^{3}[/tex] is the density of the fluid (air in this case)

[tex]V[/tex] is the velocity

[tex]A=0.17 m^{2}[/tex] is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its weight due to the gravity force [tex]W[/tex]:

[tex]W=m.g[/tex] (2)

Where:

[tex]m=79.5 kg[/tex] is the mass of the object

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]D=W[/tex] (3)

[tex]{C}_{d}\frac{\rho V^{2} }{2}A=m.g[/tex] (4)

[tex]V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}[/tex] (5) This is the terminal velocity

Substituting the known values in (5):

[tex]V=\sqrt{\frac{2(79.5 kg)(9.8 m/s^{2})}{(1.21 kg/m^{3})(0.17m^{2}){(0.7)}}[/tex] (6)

Then:

[tex]V=104.026 m/s[/tex] This is the final velocity in meters per second

Now, let's find the final velocity in kilometers per hour, knowing [tex]1 km=1000 m[/tex] and [tex]1 h=3600 s[/tex]:

[tex]V=104.026 \frac{m}{s} (\frac{1 km}{1000 m})(\frac{3600 s}{1 h})=374.49 km/h[/tex] This is the final velocity in kilometers per hour.