Terminal Velocity Calculator

This terminal velocity calculator finds the speed at which air drag and weight are equal. Example: 80 kg, a drag coefficient of 1.0 and 0.7 m² give about 42.78 m/s (154 km/h).

Example: Mass m 80 kg · Drag coefficient C_d 1 · Cross-sectional area A 0.7 m² · Air density ρ 1.225 kg/m³ · Gravitational acceleration g 9.8067 m/s² → Result: Terminal velocity v_t = 42.776 m/s (153.99 km/h). Source: Halliday/Resnick/Walker, Fundamentals of Physics, Kap. „Drag force and terminal speed“ (siehe Wikipedia – Terminal velocity). Updated: .

kg
Rule of thumb depending on shape and posture; a person falling is roughly 0.7 to 1.3.
m²
Area perpendicular to the direction of motion.
kg/m³
Standard: 1.225 kg/m³ (sea level, 15 °C). It is lower at altitude.
m/s²
Standard value 9.80665 m/s² (exact, CGPM 1901).
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Result

Result
Terminal velocity v_t = 42.776 m/s (153.99 km/h)
Terminal velocity and fall progress
QuantityValue
Terminal velocity v_t42.776 m/s
in kilometres per hour153.99 km/h
in miles per hour95.688 mph
Time to 95% of v_t7.9902 s
Distance fallen to 95% of v_t217.18 m
Calculation steps
  • At equilibrium the weight equals the drag force: m·g = ½·ρ·C_d·A·v².
  • v_t = √(2·m·g / (ρ·C_d·A)) = √(2·80·9.8067 / (1.225·1·0.7)) = 42.776 m/s = 153.99 km/h = 95.688 mph.
  • With v(t) = v_t·tanh(g·t / v_t), 95% of v_t is reached at t = artanh(0.95)·v_t / g = 7.9902 s; by then the body has fallen about 217.18 m.
Model with quadratic air drag, constant air density and no buoyancy. Real values depend on posture, altitude and weather; rule-of-thumb figures, not a safety calculation.

How it is calculated

Terminal velocity is the speed at which the drag force equals the weight, so the body stops accelerating. A falling body first speeds up, but air drag grows with the square of speed until the two forces cancel.

Formula

From m·g = ½·ρ·C_d·A·v² we get v_t = √(2·m·g / (ρ·C_d·A)). Heavier, smaller and more streamlined means faster; denser air slows more. Buoyancy of the displaced air is neglected. The time to 95% of terminal velocity follows from v(t) = v_t · tanh(g·t / v_t). The air density is the standard value of 1.225 kg/m³ from the International Standard Atmosphere.

Worked example

Mass 80 kg, C_d = 1.0, A = 0.7 m². Then v_t = √(2 · 80 · 9.80665 / (1.225 · 1.0 · 0.7)) ≈ 42.78 m/s, about 154 km/h or 95.7 mph. 95% of that is reached after about 7.99 s, during which the body falls roughly 217 m.

Frequently asked questions

How do you calculate terminal velocity?

With v_t = √(2·m·g / (ρ·C_d·A)). Example: 80 kg, C_d 1.0, 0.7 m² and an air density of 1.225 kg/m³ give about 42.78 m/s.

Why does a heavier body fall faster than a light one?

With the same shape and area, the heavier body needs more drag to balance its larger weight, and that only happens at a higher speed.

How quickly is terminal velocity reached?

Strictly only asymptotically. In practice, 95% is reached after about 8 seconds in the example.

Sources and legal basis

As of: · Editorial policy

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