A 72 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.

Question

Hank

Physics

3 September, 20:45

A 72 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.

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Kaylah Mathews · Answer 1

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