A72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag coefficient is 0.80. part a what is his terminal speed if he falls feet first? use ρ = 1.2 kg/m3 for the density of air at room temperature.

Formula for terminal velocity is:

Vt = √(2mg/ρACd) 
Vt = terminal velocity = ?
m = mass of the falling object = 72 kg
g = gravitational acceleration = 9.81 m/s^2
Cd = drag coefficient = 0.80
ρ = density of the fluid/gas = 1.2 kg/m^3
A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:

Vt =  √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80) 

Vt = 130.73 m/s

okay i have so many questions. but the big one here is: um wt..f is this mess? ?