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

Step-by-step 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)(\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)(\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 (m)/(s) ((1 km)/(1000 m))((3600 s)/(1 h))=374.49 km/h` This is the final velocity in kilometers per hour.