Classical-Mechanics · Unit 9 · Video 6 · Interactive Practice

The Speed Limit of Falling Objects: Terminal Velocity and the Hyperbolic-Tangent Law

IKey Formulas

Formula	Name	What it says
$F_{\text{drag}} = \tfrac{1}{2} C_D A \rho\, v^2 = \beta v^2$	Quadratic drag law	Resistance grows with speed squared
$m\dfrac{dv}{dt} = mg - \beta v^2$	Equation of motion	Newton's 2nd law, $+y$ pointing down
$v_\infty = \sqrt{\dfrac{mg}{\beta}} = \sqrt{\dfrac{2mg}{C_D A \rho}}$	Terminal velocity	Speed where drag cancels weight
$v(t) = v_\infty \tanh\!\left(\sqrt{\dfrac{\beta g}{m}}\; t\right)$	Velocity law	Rises, then flattens toward $v_\infty$

Key Insight: Terminal velocity is not a built-in property but an equilibrium — the exact speed at which the growing drag $\beta v^2$ rises to cancel the constant weight $mg$ , leaving zero acceleration.

IIVisualization 1 — Where Drag Balances Gravity

How fast must the object fall before air drag grows to exactly cancel its weight?

IIIVisualization 2 — Velocity Climbs to the Ceiling

Released from rest, how does velocity grow over time — and toward what ceiling?

Time

t

0.00

IVVisualization 3 — Why Heavier Falls Faster

Under the same drag law, why does a heavy skydiver's terminal speed dwarf a light raindrop's?

Mass

m

(kg) 0.8

💡 Real objects differ in both mass and area, so the same law $v_\infty = \sqrt{2mg/(C_D A \rho)}$ spans a huge range: a raindrop settles near $6.5$ m/s, a skydiver near $60$ m/s.

VQuiz Questions

Problem 1 · Compute a Terminal Velocity

Given: A falling object has weight $mg = 20\ \text{N}$ and drag constant $\beta = 0.05\ \text{kg/m}$ , so the drag force is $\beta v^2$ . Find its terminal velocity $v_\infty$ .

✅ Correct!

v_\infty = \sqrt{mg/\beta} = \sqrt{20/0.05} = \sqrt{400} = 20\ \text{m/s}

❌ Almost. That is

mg/\beta = 400

— but terminal velocity is the square root of that ratio.

❌ Watch the ½. The factor of

2

\sqrt{2mg/(C_D A\rho)}

is already absorbed into

\beta = \tfrac{1}{2} C_D A\rho

; use

\sqrt{mg/\beta}

, not

\sqrt{2mg/\beta}

❌ Not quite. Set

mg = \beta v_\infty^2

and solve for

v_\infty = \sqrt{mg/\beta}

Show solution

At terminal velocity the acceleration is zero, so drag balances weight:

mg = \beta v_\infty^2 \quad\Rightarrow\quad v_\infty = \sqrt{\frac{mg}{\beta}} = \sqrt{\frac{20}{0.05}} = \sqrt{400} = 20\ \text{m/s}.

Forgetting the square root gives $400$ ; double-counting the ½ already inside $\beta$ gives $\sqrt{800}\approx 28.3$ .

Problem 2 · Scaling with Mass

Given: An object falls with terminal velocity $v_\infty = 12\ \text{m/s}$ . A second object of the same size and shape (same $\beta$ ) but twice the mass is dropped. Find its terminal velocity.

✅ Correct! Since

v_\infty \propto \sqrt{m}

, doubling the mass multiplies it by

\sqrt{2}

12\sqrt{2}\approx 17\ \text{m/s}

❌ Close — check the power.

v_\infty = \sqrt{mg/\beta} \propto \sqrt{m}

, so doubling

m

scales it by

\sqrt{2}\approx 1.41

, not by

2

❌ Mass does matter. With

\beta

fixed, a heavier object needs a higher speed for drag to balance its larger weight, so

v_\infty

rises.

❌ Wrong direction. More mass means more weight to cancel, so terminal velocity increases, not decreases.

❌ Not quite. Use

v_\infty \propto \sqrt{m}

with

m \to 2m

Show solution

With $\beta$ held fixed, $v_\infty = \sqrt{mg/\beta}$ depends on mass as $v_\infty \propto \sqrt{m}$ . Doubling the mass:

v_\infty' = \sqrt{\frac{(2m)g}{\beta}} = \sqrt{2}\,\sqrt{\frac{mg}{\beta}} = \sqrt{2}\,(12) \approx 17\ \text{m/s}.

This is exactly why a dense object (more mass for the same area) settles at a higher terminal velocity than a light one of the same shape.

Problem 3 · From Formula to First Instant

Given: A ball of mass $m = 8\ \text{kg}$ falls with drag $\beta v^2$ where $\beta = 0.2\ \text{kg/m}$ ; take $g = 10\ \text{m/s}^2$ .

What is its terminal velocity?

At the instant of release ( $v = 0$ ), what is its acceleration?

✅ Correct!

v_\infty = \sqrt{80/0.2} = 20\ \text{m/s}

, and at

v = 0

the drag vanishes, so

a = g = 10\ \text{m/s}^2

❌ Check the terminal velocity.

v_\infty = \sqrt{mg/\beta} = \sqrt{(8)(10)/0.2} = \sqrt{400}

— remember the square root.

❌ Check the first instant. At

v = 0

the drag

\beta v^2

is zero, so the only force acting is gravity:

a = g

Show solution

Terminal velocity:

v_\infty = \sqrt{\frac{mg}{\beta}} = \sqrt{\frac{(8)(10)}{0.2}} = \sqrt{400} = 20\ \text{m/s}.

Release instant: from $m\,\dfrac{dv}{dt} = mg - \beta v^2$ , dividing by $m$ gives $a = g - \dfrac{\beta v^2}{m}$ . At $v = 0$ the drag term drops out:

a = g - \frac{\beta (0)^2}{m} = g = 10\ \text{m/s}^2.

Acceleration starts at its maximum $g$ and decays to zero as $v \to v_\infty$ .

Problem 4 · Raindrop vs. Skydiver (Transfer)

Given: A raindrop ( $v_\infty \approx 6.5\ \text{m/s}$ ) and a skydiver ( $v_\infty \approx 60\ \text{m/s}$ ) are both released from rest and obey $v(t) = v_\infty\tanh\!\left(\sqrt{\beta g/m}\; t\right)$ . Which statement is correct?

✅ Correct! At

v = 0

drag is zero for both, so both start at

a = g

; mass sets only the height of the ceiling, not the initial acceleration.

❌ Common trap. Right at release

v = 0

, so drag is zero for both and each starts at exactly

a = g

— the extra mass raises

v_\infty

, it does not change the initial acceleration.

❌ Not quite.

\tanh

approaches

1

asymptotically, so both approach their own

v_\infty

without ever exceeding it — the raindrop's ceiling is simply lower.

❌ Not quite. Faster motion means more drag (

\beta v^2

grows with

v

) — that growth is precisely what caps the speed.

Show solution

At the instant of release $v = 0$ , so $\beta v^2 = 0$ and the equation of motion gives $a = g$ for every falling object, regardless of mass or size.

Mass and area enter only through $v_\infty = \sqrt{2mg/(C_D A\rho)}$ : the skydiver's much larger mass gives a much higher ceiling, so it accelerates for longer before drag catches up. Both curves share the same $\tanh$ shape, approaching — but never exceeding — their own $v_\infty$ .

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