Terminal Velocity Calculator (Drag Equation)
Find the terminal velocity of a falling object from its mass, cross-section, drag coefficient and the fluid density.
How it works
v_t = √(2·m·g ÷ (ρ·A·C_d))
Terminal velocity is the steady speed at which the upward drag force exactly balances weight, so the object stops accelerating. Because drag rises with the square of speed, the balance point is v_t = √(2mg / ρA C_d). Heavier or more compact objects fall faster; a larger frontal area, a bluffer shape (higher C_d) or denser air all slow the descent — which is exactly how a parachute works by multiplying A.
Worked example
A 80 kg skydiver with 0.7 m² frontal area and C_d = 1.0 in sea-level air (ρ = 1.225 kg/m³): v_t = √(2 × 80 × 9.81 / (1.225 × 0.7 × 1.0)) = √(1569.6 / 0.8575) = √1830 ≈ 42.8 m/s, about 154 km/h — the classic belly-to-earth free-fall speed.
Sources: Terminal velocity — Wikipedia
Frequently asked questions
Why do heavy objects reach a higher terminal velocity?
More mass means more weight to balance, so drag must grow — and drag only grows with speed. The object therefore has to fall faster before the forces balance, giving a higher terminal velocity (it scales with the square root of mass).
How does a parachute slow you down?
It hugely increases the frontal area A (and the drag coefficient), and terminal velocity falls as 1/√A. A canopy can cut a skydiver’s ~55 m/s free-fall speed to a survivable ~5 m/s.
How is this different from Stokes’ terminal velocity?
Stokes’ law applies to tiny, slow objects in a viscous fluid where drag is linear in speed. This quadratic-drag equation applies to larger, faster objects such as skydivers, hail and dropped balls where drag scales with speed squared.
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This calculator is for educational purposes. Double-check important results before acting on them.