Question

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the terminal velocity (in meters per second and kilometers per hour) of an 80.0-kg skydiver falling in a pike (headfirst) position with a surface area of 0.140 m2

Answer 1

Answer:

115 m/s, 414 km/hr

Explanation:

There are two forces acting on a skydiver: gravity and air resistance (drag).  At terminal velocity, the two forces are equal and opposite.

∑F = ma

D − mg = 0

D = mg

Drag force is defined as:

D = ½ ρ v² C A

where ρ is the fluid density,

v is the velocity,

C is the drag coefficient,

and A is the cross sectional surface area.

Substituting and solving for v:

½ ρ v² C A = mg

v² = 2mg / (ρCA)

v = √(2mg / (ρCA))

We're given values for m and A, and we know the value of g.  We need to look up ρ and C.

Density of air depends on pressure and temperature (which vary with elevation), but we can estimate ρ ≈ 1.21 kg/m³.

For a skydiver falling headfirst, C ≈ 0.7.

Substituting all values:

v = √(2 × 80.0 kg × 9.8 m/s² / (1.21 kg/m³ × 0.7 × 0.140 m²))

v = 115 m/s

v = 115 m/s × (1 km / 1000 m) × (3600 s / hr)

v = 414 km/hr