Terminal Velocity
Constant fall speed when drag balances gravity.
Formula
Variables
Example
Skydiver ~55 m/s
How it works
This formula gives the speed at which a falling object stops speeding up because the upward drag force exactly matches its weight. It comes from setting the gravitational force mg equal to the aerodynamic drag force, which for many objects is modeled as (1/2)\rho A C_d v^2, then solving for v. The result shows that terminal speed increases with mass and decreases if the object has a larger cross-sectional area, higher drag coefficient, or is falling through denser air.
When to use it
Use this when analyzing objects falling through a fluid, especially air, after they have had time to accelerate and drag has become dominant. It is common in mechanics, skydiving, ballistics, and any problem asking for the steady-speed limit under quadratic drag.
Worked example, step by step
A small metal sphere has mass m = 0.80 kg, cross-sectional area A = 0.012 m^2, drag coefficient C_d = 0.47, and falls in air of density ρ = 1.20 kg/m^3. Use v_t = sqrt(2mg / (ρAC_d)) Substitute: v_t = sqrt((2 × 0.80 × 9.81) / (1.20 × 0.012 × 0.47)) Compute numerator: 2 × 0.80 × 9.81 = 15.696 Compute denominator: 1.20 × 0.012 × 0.47 = 0.006768 Divide: 15.696 / 0.006768 ≈ 2319.9 Square root: v_t ≈ 48.2 m/s Final answer: the terminal velocity is about 48 m/s.
Common mistakes to avoid
- Using the formula before drag is large enough to matter, rather than after the object has reached steady fall speed
- Forgetting that A is the projected frontal area, not the total surface area
- Mixing up the drag coefficient and air density, or using inconsistent units for mass, area, and density
Did You Know?
A falling human reaches a terminal velocity of about 200 km/h, but a spread-eagled skydiver falls much slower.
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