What’s the terminal velocity?
Type the mass, the frontal area and the drag coefficient of a falling object, and the air density. The terminal velocity calculator gives the speed where air drag balances the weight, so the fall stops speeding up.
- Terminal velocity
- 99.2086 mph
The terminal velocity is 99.2086 mph (44.3502 m/s).
- Terminal velocity (m/s)
- 44.3502
- Terminal velocity (mph)
- 99.2086
- Terminal velocity (km/h)
- 159.661
- Drag at terminal velocity (N)
- 833
Terminal velocity: 99.2086 mph. The terminal velocity is 99.2086 mph (44.3502 m/s).
How to calculate
Computes the terminal velocity of a falling object, v = √(2mg ÷ ρCA), from its mass, frontal area and drag coefficient and the air density, with the drag force at that speed.
Example with the default inputs (Mass 187.392922857146 lb, Frontal area 7.5347372916968 ft², Drag coefficient 1, Air density 1.21 kg/m³, Gravity (m/s²) 9.8): The terminal velocity is 99.2086 mph (44.3502 m/s).
Method: v_T = √(2 m g ÷ (ρ C A)), where the drag ½ C ρ A v² equals the weight m g.
- Drag grows with the square of the speed (a fast, turbulent flow), with a fixed drag coefficient and frontal area.
- The air density is the same along the fall; real air thins with height, so falls from very high reach higher speeds up top.
- Exact unit sizes: 1 lb = 0.45359237 kg; 1 ft² = 0.09290304 m²; 1 in² = 0.00064516 m²; 1 lb/ft³ = 0.45359237 ÷ 0.028316846592 kg/m³; 1 mph = 0.44704 m/s.
Worked examples
Each example is checked against the calculator on every build.
- Mass 187.4 lb, Frontal area 7.535 ft², Drag coefficient 1, Air density 1.21 kg/m³, Gravity (m/s²) 9.8 gives Terminal velocity (m/s) 44.350222.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½ C ρ A v²; v_T = √(2mg ÷ ρCA); Example 6.17), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (Example 6.17: 44 m/s)
- Mass 3.527 lb, Frontal area 5.382 ft², Drag coefficient 0.8, Air density 1.25 kg/m³, Gravity (m/s²) 10 gives Terminal velocity (m/s) 8, Drag at terminal velocity (N) 16.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½ C ρ A v²; v_T = √(2mg ÷ ρCA); Example 6.17), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed
- Mass 0.0009921 lb, Frontal area 0.000338 ft², Drag coefficient 0.45, Air density 1.225 kg/m³, Gravity (m/s²) 9.80665 gives Terminal velocity (m/s) 22.580963.Source: OpenStax, University Physics Volume 1, §6.4 Drag Force and Terminal Speed (F_D = ½ C ρ A v²; v_T = √(2mg ÷ ρCA); Example 6.17), https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (sphere C = 0.45)
How it works
At terminal velocity the drag force equals the weight:
½ × C × ρ × A × v² = m × g, so v_T = √(2 × m × g ÷ (ρ × C × A))
with the mass m in kg, the gravity g in m/s² (9.8 by default, as in OpenStax; 9.80665 is standard gravity), the air density ρ in kg/m³, the drag coefficient C, and the frontal area A in m². The page shows v_T in mph (÷ 0.44704), km/h (× 3.6) and m/s, and the drag force at that speed, m × g, in newtons.
Unit sizes: 1 g = 0.001 kg; 1 lb = 0.45359237 kg; 1 oz = 0.028349523125 kg; 1 cm² = 10⁻⁴ m²; 1 ft² = 0.09290304 m²; 1 in² = 0.00064516 m²; 1 lb/ft³ = 0.45359237 ÷ 0.028316846592 kg/m³.
Exact arithmetic. Each value is read as the exact decimal you typed, in its unit, so 2mg ÷ (ρCA) is an exact fraction; the square root is exact when it can be (√64 = 8) and a 64-bit float otherwise.
Output format. Every value shows 6 significant figures. The headline speed shows in mph (km/h in Metric).
Assumptions
- Drag grows with the square of the speed, with a fixed drag coefficient and frontal area, and the same air density all the way down.
- Mass 10⁻⁹ to 10⁹ kg; area 10⁻⁹ to 10⁶ m²; drag coefficient 0.001 to 10; air density 0.0001 to 20,000 kg/m³; gravity 0.001 to 1,000 m/s².
Worked examples by hand
An 85 kg skydiver spread-eagle (OpenStax Example 6.17). v_T = √(2 × 85 × 9.8 ÷ (1.21 × 1.0 × 0.70)) = √(1,666 ÷ 0.847) = √1,966.94 = 44.3502 m/s (99.2 mph).
1.6 kg, 0.5 m², C = 0.8, ρ = 1.25, g = 10. v_T = √(2 × 1.6 × 10 ÷ (1.25 × 0.8 × 0.5)) = √(32 ÷ 0.5) = √64 = 8 m/s; drag 1.6 × 10 = 16 N.
A 0.45 g sphere, 31.4 mm² (0.314 cm²) cross-section (C = 0.45, ρ = 1.225, g = 9.80665). v_T = √(2 × 0.00045 × 9.80665 ÷ (1.225 × 0.45 × 0.0000314)) = √509.90 = 22.5810 m/s.
Other questions people ask
What is the formula for terminal velocity?
v_T = √(2 m g ÷ (ρ C A)), where m is the mass, g the gravity, ρ the air density, C the drag coefficient and A the frontal area. It comes from setting the drag ½ C ρ A v² equal to the weight m g.
What is a skydiver’s terminal velocity?
For an 85 kg skydiver spread-eagle (C = 1.0, A = 0.70 m², ρ = 1.21 kg/m³), v_T = √(2 × 85 × 9.8 ÷ (1.21 × 0.70)) = 44 m/s, about 99 mph (OpenStax Example 6.17). Diving head first cuts the area and drag and raises the speed a lot.
What drag coefficient should I use?
OpenStax lists about 0.05 for an airfoil, 0.45 for a sphere, 0.70 for a skydiver feet first, 1.0 for a skydiver spread-eagle and 1.12 for a flat circular plate facing the flow.
Does a heavier object fall faster?
In air, yes: terminal velocity grows with the square root of the mass for the same shape and size. Four times the mass doubles the terminal velocity.
What air density should I use?
About 1.225 kg/m³ at sea level and 15 °C, and 1.21 kg/m³ near 20 °C. The air thins with height: at 3,000 m it is about 0.91 kg/m³, so the terminal velocity there is about 16% higher.
Why does the page give a drag force?
At terminal velocity the drag force equals the weight, m g. The page shows it in newtons as a check.