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Terminal Velocity Calculator

Estimate the steady falling speed where drag balances weight. Enter mass, drag coefficient, frontal area, fluid density, and gravity to check whether a terminal-speed scenario is plausible.

Active equation
Examples

80 kg, Cd 1.0, frontal area 0.7 m², air density 1.225 kg/m³. Terminal velocity is about 42.8 m/s.

Terminal Velocity
42.7836 m/s
Drag Force at Terminal Speed
784.8 N
Terminal Velocity (km/h)
154.021 km/h
Terminal Velocity (mph)
95.7042 mph
Terminal Velocity (ft/s)
140.3662 ft/s
Drag Force (lbf)
176.4301 lbf

Steady-state approximation only. This assumes constant drag coefficient and frontal area and solves the equilibrium point where drag balances weight. It does not model changing body position, altitude or weather changes, compressibility, or parachute deployment.

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Examples

How It Works

Formula

vt=2mgρCdAv_t = \sqrt{\frac{2 m g}{\rho C_d A}}

Fd=12ρCdAvt2=mgF_d = \frac{1}{2} \rho C_d A v_t^2 = m g

Variables

vtv_t

Terminal velocity(m/s)

FdF_d

Drag force at terminal speed(N)

mm

Mass(kg)

gg

Gravitational acceleration(m/s²)

ρ\rho

Fluid density(kg/m³)

CdC_d

Drag coefficient

AA

Frontal area(m²)

Enter the object mass, drag coefficient, frontal area, fluid density, and gravity. The calculator converts the chosen units into one SI basis, solves vt = sqrt((2mg)/(rho Cd A)), then reports the steady-state speed in m/s together with readable speed conversions and the matching drag force.

Terminal velocity comes from balancing the quadratic drag equation against weight. In steady state, drag force Fd = 1/2 rho Cd A v² matches weight mg. Solving that equality for v gives vt = sqrt((2mg)/(rho Cd A)). The calculator uses your chosen mass, area, density, and gravity units, converts them into kilograms, square metres, kilograms per cubic metre, and metres per second squared, solves the equilibrium speed, and then derives the drag force as mg at that same balance point.

Frequently Asked Questions

01What is terminal velocity?
Terminal velocity is the steady falling speed where drag force has risen enough to balance weight. At that point the net force is zero, so the object stops accelerating and continues at roughly constant speed.
02Why do heavier objects often have a higher terminal velocity?
In this formula, mass raises the numerator while drag coefficient, frontal area, and fluid density stay in the denominator. If shape and area do not change, more mass means the object needs a higher speed before drag grows enough to match weight.
03How is this different from projectile motion or speed-distance-time?
This calculator solves the equilibrium drag problem directly. Projectile motion starts from a launch speed and follows a path, while speed-distance-time only relates motion variables once a speed is already known.
04Does the calculator model changing posture, weather, or parachutes?
No. It is a steady-state approximation with constant drag coefficient and frontal area. It does not model changing body position, altitude or weather changes, compressibility, or parachute deployment.
05Why is drag force equal to weight here?
At terminal velocity the upward drag force and downward weight are equal in magnitude, so the net force is zero. That is why the drag-force output is the equilibrium force needed to balance m × g.

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