Terminal Velocity Calculator
Calculate terminal velocity from mass, gravity, fluid density, drag coefficient, and cross-sectional area.
What Is Terminal Velocity?
Terminal velocity is the constant speed a falling object reaches when the upward drag force balances its weight. Before that point, gravity accelerates the object, but air resistance grows with speed until no net acceleration remains. Skydivers, raindrops, and pollen grains all settle at characteristic terminal speeds based on their size, shape, and the surrounding fluid.
Terminal Velocity Formula
At terminal speed, drag equals weight. Solving for velocity gives:
$$v_t = \sqrt{\frac{2mg}{\rho A C_d}}$$Here $m$ is mass, $g$ is gravitational acceleration, $\rho$ is fluid density, $A$ is cross-sectional area, and $C_d$ is the drag coefficient. Sea-level air density is about 1.225 kg/m³. A human in a spread position often uses $C_d \approx 1.2$ and $A \approx 0.7$ m².
Related tools: Free Fall Calculator, Stokes' Law Calculator, and G-Force Calculator.
Frequently Asked Questions
What factors increase terminal velocity?
Heavier mass, smaller cross-sectional area, lower drag coefficient, and lower fluid density all raise terminal velocity. A compact, dense object falls faster than a light, wide one in the same air.
Why does drag depend on cross-sectional area?
Drag pushes against the surface that faces the flow. A larger frontal area intercepts more air molecules, so resistance increases. That is why skydivers slow down when they spread their arms and legs.
What is a typical drag coefficient?
A smooth sphere is near 0.47, a cube about 1.05, and a human body in a belly-to-earth position about 1.2. Streamlined shapes have lower $C_d$ values because they disturb the airflow less.
Does terminal velocity work in water?
Yes. Use water density (about 1000 kg/m³) instead of air density. Higher fluid density produces much lower terminal speeds, which is why objects settle slowly in water compared with air.
Is this formula valid at very high speeds?
The square-root drag model assumes turbulent drag proportional to $v^2$. At very low Reynolds numbers, Stokes' law applies instead. For everyday falling objects in air, the standard terminal velocity equation is a good estimate.
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