# What’s the terminal velocity?

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.

- Page: https://www.acalculator.org/physics/terminal-velocity-calculator
- JSON spec: https://www.acalculator.org/physics/terminal-velocity-calculator.json
- Version: f3c275ddc57a

## Default answer

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).

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| m | Mass | The mass of the falling object. |
| A | Frontal area | The area the object shows to the air as it falls. |
| C | Drag coefficient | The drag coefficient C: about 0.45 for a sphere, 0.7 for a skydiver feet first, 1.0 spread-eagle. |
| rho | Air density | The density of the air (or other fluid): about 1.225 kg/m³ at sea level and 15 °C. |
| g | Gravity (m/s²) | The acceleration of gravity in m/s²: 9.80665 on Earth, 9.8 in many textbooks. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| speed | Terminal velocity | The top speed of the fall, where the drag equals the weight. |
| mps | Terminal velocity (m/s) | v_T in meters per second. |
| mph | Terminal velocity (mph) | v_T in miles per hour. |
| kmh | Terminal velocity (km/h) | v_T in kilometers per hour. |
| weight | Drag at terminal velocity (N) | The drag force at that speed, equal to the weight m g, in newtons. |

## Method

v_T = √(2 m g ÷ (ρ C A)), where the drag ½ C ρ A v² equals the weight m g.

## Assumptions

- 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

1. m = 85, A = 0.7, C = 1, rho = 1.21, g = 9.8 gives mps = 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).
2. m = 1.6, A = 0.5, C = 0.8, rho = 1.25, g = 10 gives mps = 8, weight = 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.
3. m = 0.00045, A = 0.000031, C = 0.45, rho = 1.225, g = 9.80665 gives mps = 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).

## FAQ

### 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.

## Sources

- 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; Table 6.2 drag coefficients). https://openstax.org/books/university-physics-volume-1/pages/6-4-drag-force-and-terminal-speed (retrieved 2026-10-03)
- NIST SP 811, Appendix B.8 (1 lb = 0.45359237 kg; 1 ft = 0.3048 m; 1 mi = 1,609.344 m). https://www.nist.gov/pml/special-publication-811/nist-guide-si-appendix-b-conversion-factors/nist-guide-si-appendix-b8 (retrieved 2026-10-03)