Question

A 72kg skydiver can be modeled as a rectangular "box" with dimensions 18cm

Answer 1

D = p * Cd * A * v^2 * 0.5

p (rho) = 1 kg/m^3 (assumption from air)

Cd = 0.8

A = 0.18 * 0.47 = 0.0846m^2 (since he is falling feet first)

W = D = 9.8 * 72kg = 705.6N

705.6 N = (1 kg/m^3)(0.8)(.0846m^2)(0.5)V^2

Vt = 144.35 m/sec

Answer 2

Vt = sqrt (2 x m x g / ? x A x Cd)
where
Vt = terminal velocity,
m = mass of the falling object,
g = acceleration due to gravity,
Cd = drag coefficient,
? = density of the fluid through which the object is falling, and
A = projected area of the object.

V_t = sqrt (2 x 72 x 9.8 / 1.2 x 0.18 x 0.47 x 0.8)
V_t = 131.82 m/s

Answer 3

D = p * Cd * A * v^2 * .5

p (rho) = 1 kg/m^3 (assumption from air)
Cd = 0.80 (another assumption)
A = .18m * .47m *0.17= .0.014382m^2 (since he is falling feet first)
W = D = 9.8 * 72kg = 705.6N

705.6*2/1*0.80*.0.014382m^2 =v^2

705.6 /0.011=v^2

v=253.2m/sec

Answer 4

The terminal velocity is

v = sqrt(2mg/(rho * C *A))

So we need the density of air in kg/m^3 and to calculate A in m^2

For density of air (rho) I'll use 1.2 kg/m*2, but you use whatever is in your book.
I am picking the density at 20 degrees C because they don't say the actual temperature and thats about 60 F

A = 0.18 m * 0.47 m = 0.0846 m^2

v = sqrt(2 * 72 kg * 9.81 m/s^2 / (1.2 kg/m^2 * 0.0846 m^2 * 0.80))

v = 131.88 m/s

The terminal velocity is 131.88 m/s

Answer 5

D = p * Cd * A * v^2 * .5

p (rho) = 1 kg/m^3 (assumption from air)
Cd = 1 (another assumption)
A = .18m * .47m = .0846m^2 (since he is falling feet first)
W = D = 9.8 * 72kg = 705.6N

so
705.6N = (1 kg/m^3)(1)(.0846m^2)(.5)V^2
Vt = 129.1 = 129 m/s