When an object falls through a fluid, gravity accelerates it downward while air resistance pushes back up. The drag force increases as the object speeds up, and at some point the two forces balance exactly. From that moment, the object stops accelerating and falls at a constant speed called the terminal velocity. The formula for terminal velocity is the square root of two times mass times gravity divided by the product of fluid density, drag coefficient, and cross-sectional area: vt equals the square root of (2mg divided by rho times Cd times A). Each factor plays a clear physical role. A heavier object needs more drag to slow it down, so it reaches a higher terminal velocity. A larger cross-sectional area creates more drag at a given speed, so it lowers terminal velocity. A higher drag coefficient, which reflects a blunter or less streamlined shape, also lowers terminal velocity. Higher air density increases drag and reduces terminal velocity, which is why terminal velocities are lower at sea level than at altitude. A skydiver in a belly-to-earth position presents a large area (roughly 0.9 m2) with a drag coefficient of about 1.0 and reaches about 53 m/s (190 km/h). Pulling into a head-down dive reduces area to around 0.1 m2, raising speed to well over 90 m/s. This calculator accepts mass, drag coefficient, cross-sectional area, and fluid density as inputs, with presets for common scenarios. It returns terminal velocity in m/s and km/h, drag force at terminal velocity (which equals weight), and power dissipated by drag at terminal speed.
Air density at sea level and 15 °C is approximately 1.225 kg/m³. Drag coefficient varies with shape, surface roughness, and Reynolds number. Results are for estimation only.
Terminal velocity: vt = √(2mg / (ρ × Cd × A)). At terminal velocity, drag equals weight: Fdrag = 0.5 × ρ × vt² × Cd × A = mg. Power dissipated at vt: P = Fdrag × vt = mg × vt in watts. Converting: km/h = m/s × 3.6; mph = m/s × 2.237.
A skydiver has mass m = 80 kg, drag coefficient Cd = 1.0, and cross-sectional area A = 0.9 m². Air density at sea level is ρ = 1.225 kg/m³, and g = 9.81 m/s². Terminal velocity vt = √(2 × 80 × 9.81 / (1.225 × 1.0 × 0.9)) = √(1569.6 / 1.1025) = √(1423.67) = 37.73 m/s (135.8 km/h). Drag force = mg = 80 × 9.81 = 784.8 N. Power = 784.8 × 37.73 = 29,612 W. These match the default values pre-filled above.
Terminal velocity is the constant speed that a falling object reaches when the gravitational force pulling it down equals the aerodynamic drag force pushing up. At this speed, acceleration becomes zero and the object falls at a steady rate. The formula is vt = square root of (2mg divided by rho times Cd times A), where m is mass, g is gravitational acceleration, rho is fluid (air) density, Cd is the drag coefficient, and A is the cross-sectional area presented to the flow.
A skydiver in a stable, belly-to-earth position typically reaches a terminal velocity of about 53 to 60 m/s (190 to 215 km/h or 120 to 135 mph). In a head-down dive position, where the cross-sectional area is much smaller, speeds of over 90 m/s (325 km/h) are possible. The spread-eagle belly position presents the largest area (roughly 0.6 to 0.9 square metres) and has a drag coefficient of around 1.0, which limits speed.
Heavier objects (higher mass) reach a higher terminal velocity, assuming the same drag coefficient and cross-sectional area. This is because gravity pulls harder on a heavier object, requiring more drag force (and therefore more speed) to balance it. This is why a lead ball reaches the ground before a feather in air: the lead ball has a much higher weight-to-drag ratio. In a vacuum, both would fall at the same rate.
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