At higher relative speeds, many bodies moving through a fluid are approximated by a drag force whose magnitude is proportional to the square of speed.

A compact vector model is

$$\boxed{\mathbf F_d=-c|\mathbf v|\mathbf v},$$

where $c>0$ is a drag parameter determined by the body and surrounding fluid. The factor $|\mathbf v|\mathbf v$ gives magnitude proportional to $v^2$ while keeping the force opposite the velocity.

Falling under gravity

Take downward as positive and consider downward motion with $v\ge0$. Then

$$m\frac{dv}{dt}=mg-cv^2.$$

The terminal speed occurs when the net force vanishes:

$$mg-cv_t^2=0.$$

Therefore

$$\boxed{v_t=\sqrt{\frac{mg}{c}}}.$$

Using this definition, the acceleration can be written

$$\boxed{\frac{dv}{dt}=g\left(1-\frac{v^2}{v_t^2}\right)}.$$

When $v\ll v_t$, drag is small and the acceleration is close to $g$. As $v$ approaches $v_t$, the drag approaches the weight and the acceleration approaches zero.

If $v$ is temporarily larger than $v_t$ while moving downward, then

$$1-\frac{v^2}{v_t^2}<0,$$

so the acceleration is upward and the speed decreases toward the same terminal value. Terminal speed is therefore a stable steady speed for this model.

Worked example

A falling object has

$$m=80,\mathrm{kg}$$

and quadratic drag parameter

$$c=0.25,\mathrm{kg/m}.$$

Its terminal speed is

$$v_t=\sqrt{\frac{(80)(9.81)}{0.25}} \approx56.0,\mathrm{m/s}.$$

At half that speed,

$$v=28.0,\mathrm{m/s}=\frac12v_t,$$

so the downward acceleration is

$$a=g\left(1-\frac14\right) =\frac34g \approx7.36,\mathrm{m/s^2}.$$

At terminal speed, the same expression gives $a=0$.

How changing the parameters changes terminal speed

From

$$v_t=\sqrt{\frac{mg}{c}},$$

a larger mass increases terminal speed, while stronger quadratic resistance lowers it. Because of the square root, multiplying the mass by four doubles $v_t$, while multiplying $c$ by four halves $v_t$.

Relation to aerodynamic drag models

In aerodynamic applications, the quadratic coefficient can be expressed in terms of fluid density, body area, shape, orientation, and flow regime. Those details determine the value of $c$; the mechanics here uses the resulting force law to predict the body's motion.

Quadratic drag therefore connects a fluid-force model to Newtonian particle dynamics: resistance grows rapidly with speed, producing a finite terminal speed and strongly non-constant acceleration.