Terminal Velocity Calculator

Determine the maximum velocity achievable by a falling object

Input any five values to solve for the remaining one using the terminal velocity formula.

Last updated: December 25, 2025

CreatorFrank Zhao

Shape

More info

Coefficient of drag (Cd)

More info

Mass (m)

More info

Cross-sectional area (A)

More info

m²

Density of fluid (ρ)

More info

kg/m³

Acceleration due to gravity (g)

More info

m/s²

Terminal velocity (Vₜ)

More info

m/s

1Terminal Velocity

v_t = \sqrt{\frac{2mg}{\rho A C_d}}

2Mass

m = \frac{v_t^2 \rho A C_d}{2g}

3Area & Density

A = \frac{2mg}{\rho C_d v_t^2}

\rho = \frac{2mg}{A C_d v_t^2}

4Gravity

g = \frac{\rho A C_d v_t^2}{2m}

vₜTerminal Velocity

mMass

gGravity

ρFluid Density

ACross-section

CdDrag Coeff.

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What is terminal velocity?

Terminal velocity is the steady top speed an object reaches while falling through a fluid (like air or water). At first, gravity wins and the object speeds up. As speed increases, drag increases too — until the forces balance.

✅ When the downward weight and the upward drag are equal, the net force is zero, so the acceleration becomes $0$ — the speed stops increasing.

The classic mental picture is skydiving: a person accelerates for a while, then settles into a nearly constant falling speed. But the same idea applies to a baseball, a raindrop, a coin, or even a sinking object in water.

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Factors affecting terminal velocity and examples

In this calculator, terminal velocity depends on five key inputs: mass, cross-sectional area, drag coefficient, fluid density, and gravity. You can think of them as two buckets: object properties and environment properties.

Object-dependent

Drag coefficient ( $C_d$ ): mainly shape-related. Sleek shapes generally have lower drag than blunt ones.
Area ( $A$ ): bigger projected area usually means more drag, so a lower terminal speed.
Mass ( $m$ ): heavier objects tend to have higher terminal velocity (all else equal).

Environment-dependent

Fluid density ( $\rho$ ): denser fluids (water vs air) increase drag, reducing terminal velocity.
Gravity ( $g$ ): higher gravity increases weight, which pushes terminal velocity upward.

A practical rule of thumb

Lower $m$ with bigger $A$ often means a much lower terminal speed — that’s basically the idea behind a parachute. Same person, same mass, but a dramatically larger area and higher effective drag.

🧠 Quick check: at terminal velocity, acceleration is $0$ and the speed is (approximately) constant.

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How to calculate terminal velocity

The calculator uses the standard drag-force balance model. When drag equals weight, the net force is zero:

Force balance (at terminal speed)

mg = \tfrac{1}{2}\,\rho\,v_t^2\,A\,C_d

Where $m$ is mass, $g$ is gravitational acceleration, $\rho$ is fluid density, $A$ is cross-sectional area, $C_d$ is drag coefficient, and $v_t$ is terminal velocity.

Solving that for terminal velocity gives:

Terminal velocity formula

v_t = \sqrt{\frac{2mg}{\rho\,A\,C_d}}

Pick a shape (or choose a custom drag coefficient)

Selecting a shape fills in a typical $C_d$ for that geometry. If you want your own value, choose Enter a custom drag coefficient.

Enter mass and cross-sectional area

Use $m$ and $A$ for the object you’re modeling. (If you’re not sure about area, it’s often the "front-facing" area as it falls.)

Confirm the environment (density and gravity)

The default density is for air near room temperature, and the default $g$ is Earth’s gravity. Change them for different altitudes, fluids, or planets.

Read the result

The calculator returns $v_t$ . You can also switch units (for example, to mph) to match your situation.

💡

Tip:

If your goal is to estimate time to fall rather than the final steady speed, pair this with a free-fall calculator. Terminal velocity can be reached quickly in some cases (skydiving), and barely reached at all in others.

🧾

Example: Using the terminal velocity calculator

Suppose you want a quick estimate for a skydiver with mass $m = 75\ \mathrm{kg}$ and cross-sectional area $A = 0.18\ \mathrm{m^2}$ . Let’s use a custom drag coefficient $C_d = 0.7$ . The calculator already pre-fills air density near room temperature and Earth gravity.

Enter the inputs

Set $m = 75\ \mathrm{kg}$ , $A = 0.18\ \mathrm{m^2}$ , and choose Enter a custom drag coefficient to input $C_d = 0.7$ .

Interpret the result

The calculator outputs the terminal velocity $v_t$ in the chosen units. For this example, using $\rho = 1.2041\ \mathrm{kg/m^3}$ and $g = 9.81\ \mathrm{m/s^2}$ :

v_t

=

\sqrt{\frac{2mg}{\rho A C_d}}

=

\sqrt{\frac{2\cdot 75\cdot 9.81}{1.2041\cdot 0.18\cdot 0.7}}

=

98.48\ \mathrm{m/s}

What that number means

$98.48\ \mathrm{m/s}$ is roughly the “settled” speed once the fall has gone on long enough for drag to balance weight. In real life, body posture changes $C_d$ and effective $A$ , so treat this as a physics-based estimate.

Try it yourself: switch the output to mph and test a smaller object (like a coin) — you’ll see how strongly $m$ , $A$ , and $C_d$ drive the result.

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FAQs

What do you mean by terminal velocity?

It’s the steady speed a falling object approaches when drag grows large enough to balance weight. At that point the net force is approximately $0$ , so acceleration is approximately $0$ .

What is the terminal velocity formula?

In this model:

v_t = \sqrt{\frac{2mg}{\rho\,A\,C_d}}

Where $m$ , $g$ , $\rho$ , $A$ , and $C_d$ are the inputs.

How do I find terminal velocity step by step?

Compute the numerator: $2mg$

Multiply the mass by 2 and by gravitational acceleration.

Compute the denominator: $\rho\,A\,C_d$

Multiply the fluid density, cross-sectional area, and drag coefficient together.

Divide the two: $\dfrac{2mg}{\rho\,A\,C_d}$

Take your numerator and divide by your denominator.

Take the square root: $v_t = \sqrt{\dfrac{2mg}{\rho\,A\,C_d}}$

Apply the square root to get the terminal velocity.

What is the terminal velocity of a baseball?

It depends on how you model the ball (size, drag coefficient, etc.). Using one common set of assumptions:

v_t

=

\sqrt{\frac{2\cdot 0.14883\cdot 9.81}{1.2041\cdot 0.004393\cdot 0.3275}}

=

40.7\ \mathrm{m/s}

\approx

91.84\ \mathrm{mph}

Use this as a ballpark estimate — real-world spin, seams, and orientation can shift the effective drag.

What is the terminal velocity of a golf ball?

With one typical assumption set:

v_t

=

\sqrt{\frac{2\cdot 0.03544\cdot 9.81}{1.2041\cdot 0.001385442\cdot 0.389}}

=

32.73\ \mathrm{m/s}

The key drivers are the smaller area and a different drag coefficient compared with larger balls.

Friendly disclaimer

This calculator is a physics-based estimator built on a simplified drag model. For safety-critical planning (skydiving, ballistics, engineering), use verified data and professional guidance.

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