A 72kg skydiver can be modeled as a rectangular "box" with dimensions 18cm
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
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
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
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.
Vt = sqrt (2 x 72 x 9.8 / 1.2 x 0.18 x 0.47 x
0.8)
Vt = 131.82 m/s
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
A 72kg skydiver can be modeled as a rectangular "box" with dimensions 18cm
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