Estimate terminal velocity from mass, drag coefficient, frontal area and air density, with drag force and kinetic energy at that speed.
Spherical-drag-style model: v = √(2·m·g / (ρ·Cd·A)). This is a simplified steady-state estimate with a constant drag coefficient; real falling bodies tumble, change area and pass through changing air density.
Terminal velocity (m/s)
How to use this tool
Body mass in kilograms or pounds; the calculator converts pounds to kilograms.
Use a textbook Cd for the shape and posture (about 0.7–1.0 for a spread skydiver, lower for a streamlined body) and the projected frontal area in m².
Sea-level standard air is 1.225 kg/m³; density falls with altitude.
The headline speed is in m/s with a km/h equivalent, plus the balancing drag force and kinetic energy at that speed.
Formula & worked example
An 80 kg body with Cd = 1.0, frontal area 0.7 m² in air at 1.225 kg/m³: v = √(1569.6 / 0.8575) = 42.78 m/s (154.0 km/h), drag 784.8 N.
Limitations — read this before you rely on it
This is a simplified model: Cd is treated as constant, area is treated as fixed, and air density is treated as constant during the fall. Real skydivers and irregular objects vary all three, so treat the result as an estimate, not a safety-critical value.
Frequently asked questions
Terminal velocity grows with the square root of mass, so doubling mass raises it by about 41%, all else equal.
A larger frontal area raises drag, so the speed at which drag balances weight falls.
Roughly yes for a belly-to-earth skydiver — about 53 m/s — while a head-down skydiver can exceed 80 m/s.
Yes: thinner air at altitude gives less drag, so terminal velocity is higher until the faller reaches denser air.
No — it gives only the steady terminal speed, not the time or distance needed to reach it.