Question

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.)

Answer 1

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" $D$:  

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:  

$C_ {d}=0.7$ is the drag coefficient  

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

$V$ is the velocity  

$A=0.17 m^{2}$ 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 $W$:  

$W=m.g$ (2)

Where:  

$m=79.5 kg$ is the mass of the object

$g=9.8 m/s^{2}$ is the acceleration due gravity  

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

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

Substituting the known values in (5):

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

Then:

$V=104.026 m/s$ This is the final velocity in meters per second

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

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