Solve a quadratic-drag branch with explicit sign limits

The following quadratic-resistance and hyperbolic-function derivation is optional enrichment. The core resistive-force outcome is developed through the preceding linear-law initial-value lesson; no prior hyperbolic-function fluency is assumed.

Declare the branch before solving

Let downward be positive. For an object released from rest under gravity and quadratic drag, m dv/dt=mg−cv² applies while v is nonnegative. The general signed drag expression is−c|v|v. If velocity were upward, a formula always pointing negatively would reinforce upward motion instead of resisting it. The present closed form is therefore a rest-release downward branch, not a universal trajectory.

Separate a nonlinear differential equation

Define vt=√(mg/c). The equation becomes dv/dt=g[1−(v/vt)²]. Put u=v/vt and integrate du/(1−u²)=(g/vt)dt from zero to the current value. Since the integral is artanh(u), u=tanh(gt/vt). Hence v=vt tanh(t/tau), where tau=vt/g. This hyperbolic function is a different shape from the exponential transient of linear resistance.

Derive the other motion quantities

Differentiate to obtain acceleration g sech²(t/tau). Integrating velocity gives y=(vt²/g)ln cosh(gt/vt), with y0=0. At zero time, v and y vanish while acceleration equals g. At late times, speed approaches vt and acceleration approaches zero; displacement grows without a finite upper bound in this unlimited-height idealization.

Check dimensions and a limit

The coefficient c has units kg/m because cv² must be a force. For m=2 kg,c=.5 kg/m,g=10, vt=√40 m/s and tau=√40/10 s. Increasing c at fixed mass lowers terminal speed as c^−1/2. At very early times tanh(z)≈z, so v≈gt and y≈gt²/2, recovering the no-drag initial limit. This local limit is not permission to ignore drag at all times.

Compare evidence before choosing a law

Linear and quadratic resistance can both approach terminal speeds. One endpoint cannot identify which law holds. Compare transients over a stated range, quantify timing and speed uncertainty, and inspect whether the assumed constant coefficients remain plausible. A curve fit is evidence conditional on that range, not proof of a universal fluid law.

Transfer and safety boundary

For m=3,g=12,c=1, terminal speed is 6 m/s. Do not use this number as a safe falling speed or extrapolate the model to a physical drop. An upward launch, changing density, changing area or finite ground event requires new sign, parameter or boundary treatment. The workbench supplies a mathematical comparison only.

Public numerical transfer practice

Try a new practice case

For mass 3 kg,g=12 m/s² and c1 kg/m, find terminal speed.

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original prose, models and cases, checked 30 October 2026 against the current official AP Physics C Mechanics CED and OpenStax University Physics. Values are supplied or generated, not measured apparatus records. No physical launch, suspension, moving machinery or practical safety validation is requested. Virtual captures and public answers do not certify independent mastery or laboratory participation.