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
