Find the top speed of a falling object when air drag balances its weight, v = √(2mg/ρAC_d). Enter the mass, drag coefficient, frontal area and air density (or pick a preset) to see the terminal velocity, how long and how far it takes to reach 95% of it, and the full speed curve.
Enter the drag coefficient (or start from a preset).
Give the frontal area directly, or a diameter for round objects.
Pick the air density for your altitude or a custom fluid.
Read the terminal velocity, the time to 95% and the fall table and chart.
Terminal velocity formulas
Quantity
Formula
Drag force
Fd = ½ ρ v² Cd A
Terminal velocity
vt = √(2 m g / (ρ A Cd))
Speed after time t (from rest)
v(t) = vt tanh(g t / vt)
Distance after time t
d(t) = (vt² / g) ln cosh(g t / vt)
Time to 95% of vt
t = (vt / g) × artanh 0.95 ≈ 1.832 vt / g
Typical drag coefficients
Shape
Cd (approx.)
Flat plate facing the flow
1.1 – 1.3
Open parachute
1.3 – 1.5
Skydiver, belly-to-earth
1.0 – 1.3
Cube
1.05
Long cylinder, side-on
0.8 – 1.2
Smooth sphere
0.47
Golf ball (dimpled, at speed)
0.25 – 0.3
Streamlined body
0.04 – 0.1
Air density by altitude (standard atmosphere)
Altitude
Density (kg/m³)
Sea level
1.225
1,000 m
1.112
2,000 m
1.007
3,000 m
0.909
4,000 m
0.819
5,000 m
0.736
10,000 m
0.414
Frequently asked questions
What is the formula for terminal velocity?
v_t = √(2 m g / (ρ A C_d)), where m is mass, g gravity, ρ the fluid density, A the frontal area and C_d the drag coefficient. It is the speed at which the drag force ½ρv²C_dA equals the weight mg.
How fast does a skydiver fall?
In a stable belly-to-earth position a typical skydiver reaches roughly 50–55 m/s (about 190–200 km/h, 120 mph). Head-down positions reduce the drag area and can exceed 70 m/s (250+ km/h).
How long does it take to reach terminal velocity?
Starting from rest with quadratic drag, v(t) = v_t tanh(g t / v_t). You reach 95% of v_t after t = 1.83 v_t / g; for a 53 m/s skydiver that is about 10 seconds and roughly 340 m of fall.
Why do heavier objects fall faster in air?
Terminal velocity grows with √(m / (A C_d)). Doubling the mass of an object with the same shape and size raises its terminal speed by about 41%. In a vacuum all objects fall at the same rate.
Why is terminal velocity higher at high altitude?
Air is thinner higher up, so ρ is smaller and the same drag needs a higher speed. At 10 km the air density is about a third of sea level, so terminal velocity is about 1.7 times higher.
Is the drag coefficient constant?
Only approximately. C_d depends on shape, orientation, surface roughness and the Reynolds number, so it can change with speed (the drag crisis of balls is an example). The calculator assumes a constant C_d.
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