Question

A 90-kg skydiver jumps from a height of 6000 m above the ground, falling head-first (pike position). The area of the diver is 0.14 m^2. The value for C is 1.0, and the density of air is 1.21 kg/m^3. Determine the terminal velocity.

a. 49 m/s
b. 68 m/s
c. 83 m/s
d. 102 m/s

Answer 1

The correct option is d: 102 m/s.

Step-by-step explanation:

The terminal velocity is given by:

`v = \sqrt{(2mg)/(\rho AC)}` (1)

Where:

m: is the mass = 90 kg

g: is the gravity = 9.81 m/s²

A: is the area = 0.14 m²

C: is the drag coefficient = 1.0

ρ: is the density of the air = 1.21 kg/m³

Now, by entering the above values into equation (1) we have:

`v = \sqrt{(2mg)/(\rho AC)} = \sqrt{(2*90 kg*9.81 m/s^(2))/(1.21 kg/m^(3)*0.14 m^(2)*1.0)} = 102 m/s`

Therefore, the correct option is d: 102 m/s.

I hope it helps you!