Your weak spots

Insights load after your first practice round.

Module 16 · L10 of 16 ~45 min ⚡ +90 XP available

Projectile Motion with Air Resistance (Proportional to v²)

At higher speeds, drag scales not with $v$ but with $v^2$, the regime of cars, cannonballs, and parachutists at full speed. The ODE becomes nonlinear, but it is still separable. This lesson shows how partial fractions and $\tanh$ identities unlock $v(t)$ and terminal velocity for the horizontal and vertical cases, and why the full two-dimensional problem refuses to decouple.

Today's hook, A skydiver in free fall hits roughly $55\text{ m/s}$ at terminal velocity. Drag is $kv^2$, so the equation of motion is $m\dot v = mg - kv^2$. Before reading on, predict the terminal velocity in terms of $m, g, k$ by setting $\dot v = 0$. Why is the answer a square root this time, not a simple ratio? Compare your reasoning after card 05.

0/5QUESTS

You are here

Recall, your gut answer first

+5 XP warm-up A particle of mass moves horizontally and the only force is air resistance of magnitude (opposing motion).

Recall, your gut answer first

+5 XP warm-up

A particle of mass $m$ moves horizontally and the only force is air resistance of magnitude $kv^2$ (opposing motion). Write Newton's second law as a differential equation in $v$. Before checking separate variables and find $v(t)$ given $v(0) = v_0$. Is the decay still exponential, or something else?

auto-saved

The two moves for quadratic drag

+5 XP to read Quadratic drag breaks the linear-drag pattern.

The two moves for quadratic drag

+5 XP to read

Quadratic drag breaks the linear-drag pattern. Two habits matter: always check the sign of the velocity when writing the drag term (since $v^2$ is symmetric, you need the magnitude $|v|$ times the unit vector $-\hat v$), and use separation of variables with $v_T^2 = mg/k$ as a substitution to keep the algebra compact. The vertical case yields $\tanh$ or $\coth$ depending on whether you start below or above terminal velocity.

The sign-then-separate reading: (1) write drag as $-k v^2$ if $v > 0$ and $+k v^2$ if $v < 0$ (drag opposes motion), (2) separate $\frac{dv}{g \pm (k/m)v^2}$ and integrate using either partial fractions or an arctan/$\tanh^{-1}$, (3) read off $v_T = \sqrt{mg/k}$ as the equilibrium.

Horizontal: $\dot v = -(k/m)v^2$ · Falling: $\dot v = g - (k/m)v^2$ · Rising: $\dot v = -g - (k/m)v^2$.

Sign-and-separate flow for quadratic drag as a sequence

$m\dot v = mg - k v^2 \;\Rightarrow\; v_T = \sqrt{mg/k}$

Drag must oppose motion

Because $v^2 \geq 0$, write drag as $-kv|v|$ or split into rising / falling phases. Writing $-kv^2$ uncritically loses the sign reversal on descent and gives wrong physics.

Components do NOT decouple

In 2D the drag is $-k|\mathbf v|\mathbf v$, so the horizontal force depends on $v_y$ via $|\mathbf v| = \sqrt{v_x^2 + v_y^2}$. The general projectile problem is nonlinear and needs numerical methods.

$\tanh$ is the natural solution

For free fall from rest with quadratic drag, $v(t) = v_T \tanh(gt/v_T)$. The identity $1 - \tanh^2 = \operatorname{sech}^2$ matches the algebra. Recognising $\tanh$ saves time.

What you'll master

Know

Key facts

Quadratic drag model: force $= -k v |v|$ (or $-kv^2$ in 1D when $v > 0$)
Horizontal equation: $m\dot v = -k v^2$, solution $v = v_0 / (1 + k v_0 t / m)$
Vertical fall (from rest): $m\dot v = mg - k v^2$, solution $v = v_T \tanh(gt/v_T)$
Terminal velocity: $v_T = \sqrt{mg/k}$

Understand

Concepts

Why horizontal velocity now decays algebraically (as $1/t$), not exponentially
Why terminal velocity is a square root, not a ratio
Why the 2D problem does not decouple under quadratic drag

Can do

Skills

Apply separation of variables to $\dot v = a - bv^2$ using partial fractions or $\tanh^{-1}$
Recognise when $\tanh$, $\coth$, or $\arctan$ is the natural antiderivative
Identify and handle the sign of $v$ when writing the drag term

Key terms

Quadratic dragResistive force of magnitude $k v^2$, proportional to the square of the speed. Appropriate at higher speeds (turbulent flow), e.g., projectiles in air.

Terminal velocity (quadratic)$v_T = \sqrt{mg/k}$. The constant speed at which $mg = kv_T^2$, gravity exactly balanced by drag.

$\tanh$ solutionFor a body falling from rest, $v(t) = v_T \tanh(gt/v_T)$. Asymptotic to $v_T$ as $t \to \infty$; its derivative gives $\dot v = g \operatorname{sech}^2(gt/v_T)$.

Partial fractionsThe decomposition $\frac{1}{v_T^2 - v^2} = \frac{1}{2 v_T}\!\left(\frac{1}{v_T - v} + \frac{1}{v_T + v}\right)$, used to integrate the falling-phase ODE without quoting $\tanh^{-1}$.

Algebraic vs exponential decayLinear drag: $v$ decays as $e^{-kt/m}$. Quadratic drag (horizontal): $v$ decays as $1/(1 + ct)$, much slower at large $t$.

Non-decoupling in 2DUnder quadratic drag the force on $v_x$ depends on $\sqrt{v_x^2 + v_y^2}$, coupling the components. The full 2D problem is generally non-analytic.

MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving resisted motion of a projectile, including resistance proportional to the square of the velocity.

Horizontal component, algebraic decay

A particle of mass $m$ moves horizontally with initial speed $v_0 > 0$.

Horizontal component, algebraic decay

core concept

A particle of mass $m$ moves horizontally with initial speed $v_0 > 0$. The only force is drag $-k v^2$. Newton II gives

$$m\frac{dv}{dt} = -k v^2 \quad\Longrightarrow\quad \frac{dv}{v^2} = -\frac{k}{m}\,dt.$$

Integrating from $0$ to $t$ with $v(0) = v_0$:

$$-\frac{1}{v} + \frac{1}{v_0} = -\frac{k}{m}t \;\Longrightarrow\; v(t) = \frac{v_0}{1 + \dfrac{k v_0}{m} t}.$$

A second integration (with $x(0) = 0$) gives

$$x(t) = \frac{m}{k}\ln\!\left(1 + \frac{k v_0}{m} t\right).$$

The contrast with linear drag. Velocity decays like $1/t$ rather than $e^{-t}$, a much slower long-time decay. The displacement, however, grows like $\ln t$ and is unbounded: the particle keeps creeping forward forever, unlike the linear-drag case which had a hard limit $mu/k$.

Connecting to limits. As $t \to \infty$, $v \to 0$ and $x \to \infty$ logarithmically. Quadratic drag is "softer" than linear drag at low speeds (since $v^2 \ll v$ when $v \ll 1$), so the particle is harder to stop completely.

Horizontal ODE: $\dot v = -(k/m) v^2$, separable · $v(t) = v_0 / (1 + (k v_0/m) t)$, algebraic decay · $x(t) = (m/k)\ln(1 + (k v_0/m) t)$, logarithmic, unbounded · Decay is slower than under linear drag at large $t$

Pause, copy the quadratic-drag ODE $\dot v = -(k/m)v^2$, the algebraic-decay solution $v = v_0/(1+kv_0 t/m)$, the logarithmic position, and the contrast with linear-drag exponential decay into your book.

Quick check: A particle moves horizontally with $\dot v = -k v^2$ per unit mass and $v(0) = v_0$. Which expression gives $v(t)$?

Vertical fall & terminal velocity

We just saw horizontal quadratic drag: $\dot v = -(k/m)v^2$ gives algebraic decay $v = v_0/(1+(kv_0/m)t)$, with position growing logarithmically, distance is unbounded, unlike under linear drag.

Vertical fall & terminal velocity

core concept

We just saw horizontal quadratic drag: $\dot v = -(k/m)v^2$ gives algebraic decay $v = v_0/(1+(kv_0/m)t)$, with position growing logarithmically, distance is unbounded, unlike under linear drag. That raises a question: what is the terminal velocity under quadratic drag during vertical fall? This card answers it → $m\dot v = mg - kv^2$ gives $v_T = \sqrt{mg/k}$ (a square root, not a ratio), and $v(t) = v_T\tanh(gt/v_T)$.

Take downward as positive for the falling body of mass $m$. The forces are gravity $+mg$ and drag $-k v^2$ (drag opposes the downward motion). Newton II gives

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

Setting $\dot v = 0$ gives the terminal velocity at once: $mg = k v_T^2$, so

$$v_T = \sqrt{\frac{mg}{k}}.$$

To solve the ODE, rewrite using $v_T$ as $\dot v = (g/v_T^2)(v_T^2 - v^2)$ and separate:

$$\frac{dv}{v_T^2 - v^2} = \frac{g}{v_T^2}\,dt.$$

Use partial fractions $\frac{1}{v_T^2 - v^2} = \frac{1}{2 v_T}\!\left(\frac{1}{v_T - v} + \frac{1}{v_T + v}\right)$, integrate, apply $v(0) = 0$ and rearrange to obtain

$$v(t) = v_T \tanh\!\left(\frac{g t}{v_T}\right).$$

Why $\tanh$ shows up. The substitution $v = v_T \tanh u$ converts $v_T^2 - v^2 = v_T^2 \operatorname{sech}^2 u$ and $dv = v_T \operatorname{sech}^2 u\,du$. The separation collapses to $du = (g/v_T) dt$, integrating to $u = gt/v_T$. The hyperbolic substitution is to quadratic drag what the integrating factor was to linear drag.

Falling ODE (downward positive): $m\dot v = mg - kv^2$ · Terminal velocity: $v_T = \sqrt{mg/k}$, a square root, not a ratio · Solution from rest: $v(t) = v_T \tanh(gt/v_T)$ · Partial fractions: $1/(v_T^2 - v^2) = (1/2v_T)[1/(v_T-v) + 1/(v_T+v)]$

Pause, copy the falling ODE $m\dot v = mg-kv^2$, terminal velocity $v_T = \sqrt{mg/k}$, the tanh solution $v(t) = v_T\tanh(gt/v_T)$, and the partial-fraction split $1/(v_T^2-v^2)$ into your book.

Did you get this? True or false: for a body falling from rest under gravity with quadratic drag, the terminal velocity is $v_T = \sqrt{mg/k}$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · HORIZONTAL QUADRATIC DRAG

A particle of unit mass moves horizontally with $v(0) = 10\text{ m/s}$ and experiences air resistance $0.1 v^2$. Find $v$ as a function of $t$, and find the time at which $v = 5\text{ m/s}$.

Newton II: $\dot v = -0.1 v^2$. Separate: $dv/v^2 = -0.1\,dt$. Integrate: $-1/v = -0.1 t + C$.

For $\dot v = -kv^2$, separation gives $\int v^{-2}\,dv = -kt$, i.e., $-1/v$. Always remember the negative on the antiderivative.

PROBLEM 2 · VERTICAL FALL FROM REST

A body of mass $m$ falls from rest. Air resistance is $kv^2$ (downward positive). Find $v(t)$ in closed form using partial fractions, and confirm the terminal velocity.

Newton II: $m\dot v = mg - k v^2$. Let $v_T^2 = mg/k$. Then $\dot v = \dfrac{g}{v_T^2}(v_T^2 - v^2)$. Separate: $\dfrac{dv}{v_T^2 - v^2} = \dfrac{g}{v_T^2}\,dt$.

Introducing $v_T$ as a substitution keeps the algebra compact and makes the limit transparent. The right-hand side is now constant.

PROBLEM 3 · 2D PROJECTILE WITH QUADRATIC DRAG

A projectile is launched with velocity $(u, w)$ in air with quadratic drag of magnitude $k|\mathbf v|^2$. Write the component equations and explain why they do not decouple.

Speed: $|\mathbf v| = \sqrt{v_x^2 + v_y^2}$. Drag force: $\mathbf F_d = -k|\mathbf v|\,\mathbf v$ (so its magnitude is $k|\mathbf v|^2$ and direction opposes $\mathbf v$).

For quadratic drag we must write the vector form $-k|\mathbf v|\mathbf v$, not $-k v_x^2$ component-by-component, because the magnitude couples the components.

Fill the gap: A body falls from rest under gravity with quadratic drag $kv^2$. The terminal velocity is $v_T = $ , and the velocity at time $t$ is $v(t) = v_T \cdot $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Writing $-kv^2$ on the way down without thought

$v^2$ is symmetric in sign, but drag must oppose motion. On the descent, $v < 0$ (if upward is positive), so the drag force is $+kv^2$, pointing up. Write drag as $-k v |v|$ to get the sign automatically.

Trap 02

Decoupling components in 2D

Under quadratic drag, the horizontal force is $-k|\mathbf v|v_x$, it depends on $v_y$ through $|\mathbf v|$. You cannot solve the horizontal ODE without knowing the vertical motion. HSC restricts to 1D for this reason.

Trap 03

Quoting $v_T = mg/k$ for quadratic drag

That formula belongs to linear drag. For quadratic drag, equilibrium $mg = k v_T^2$ gives $v_T = \sqrt{mg/k}$. Mixing the two costs marks even when the structure of the rest of the solution is right.

Did you get this? True or false: under quadratic air resistance, the horizontal and vertical components of a 2D projectile's velocity satisfy independent ODEs that can be solved separately.

Revisit your thinking

Activities · practice with the ideas

A particle of unit mass moves horizontally with $v(0) = 20\text{ m/s}$ and air resistance $0.05 v^2$. Find $v(t)$ and the time at which $v = 4\text{ m/s}$.

A body of mass $m = 1\text{ kg}$ falls from rest with air resistance $0.04 v^2$. Take $g = 10\text{ m/s}^2$. Find the terminal velocity. Estimate when $v$ reaches $90\%$ of $v_T$.

For the body in question 2, derive an expression for $v$ as a function of distance fallen $y$ using $v\,dv/dy = g - (k/m) v^2$.

A particle is projected vertically upward with speed $u$ in a medium of quadratic resistance $k v^2$ per unit mass. Write the equation of motion for the ascent and show that the time to reach the highest point is $T = \frac{1}{\sqrt{gk}} \arctan(u \sqrt{k/g})$.

Compare the long-time behaviour of horizontal velocity under linear vs quadratic drag. Why is quadratic drag "softer" at small $v$?

Odd one out: Three of these statements about quadratic air resistance ($\text{drag} = kv^2$) are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted the terminal velocity for a skydiver under quadratic drag by setting $\dot v = 0$ in $m\dot v = mg - kv^2$.

The answer is $v_T = \sqrt{mg/k}$, a square root, because $v$ enters the equation squared. The full solution $v(t) = v_T \tanh(gt/v_T)$ approaches $v_T$ as $t \to \infty$, but more slowly than the exponential approach of the linear-drag case at short times, faster at long times. The structural lesson is that the shape of the drag law dictates the shape of the antiderivative: linear drag gives exponentials, quadratic drag gives logs and hyperbolics. Recognising the algebraic signature of each is what makes Module 16 efficient.

auto-saved

Show what you have learned

Multiple choice, then short answer under exam conditions.

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. A particle moves horizontally with initial speed $v_0$ and air resistance $kv^2$ per unit mass. Derive $v(t) = v_0/(1 + k v_0 t)$. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. A body of mass $m$ falls from rest under gravity with air resistance $kv^2$ (downward positive). Show that $v(t) = v_T \tanh(gt/v_T)$, where $v_T = \sqrt{mg/k}$. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A particle is projected vertically upward with speed $u$ in a medium of resistance $kv^2$ per unit mass. (a) Write the equation of motion for the ascent. (b) Find the maximum height reached. (c) State (without solving) why the descent obeys a different ODE. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $\dot v = -0.05 v^2 \Rightarrow 1/v = 0.05 t + 1/20 \Rightarrow v(t) = 20/(1+t)$. Set $v = 4$: $1+t = 5$, so $t = 4\text{ s}$.

2. $v_T = \sqrt{mg/k} = \sqrt{10/0.04} = \sqrt{250} \approx 15.81\text{ m/s}$. $v(t) = v_T \tanh(g t / v_T)$; $\tanh = 0.9$ at argument $\operatorname{arctanh}(0.9) = \tfrac12 \ln 19 \approx 1.472$, so $t \approx 1.472 \cdot v_T / g \approx 2.33\text{ s}$.

3. $v\frac{dv}{dy} = g - (k/m)v^2$. Let $u = v^2$, so $du/2 = v\,dv$. Equation becomes $\frac{m}{2k}\int \frac{du}{v_T^2 - u} = \int dy$, giving $-\frac{m}{2k}\ln(v_T^2 - u) = y + C$. $v(0)=0$: $C = -\frac{m}{2k}\ln(v_T^2)$. So $v^2 = v_T^2(1 - e^{-2 k y/m})$.

4. Ascent (upward positive): $\dot v = -g - kv^2 = -(g + kv^2)$. Separate $\frac{dv}{g + kv^2} = -dt$. Sub $v = \sqrt{g/k}\tan\theta$: $\frac{1}{\sqrt{gk}}d\theta = -dt$. Apply $v(0)=u$, $v(T)=0$: $T = \frac{1}{\sqrt{gk}}\arctan\!\left(u\sqrt{k/g}\right)$.

5. Linear drag: $v(t) = u e^{-kt/m}$, exponential, fast decay. Quadratic drag: $v(t) = u/(1 + ku t/m)$, algebraic $\sim 1/t$. At small $v$, $v^2 \ll v$, so the drag force becomes negligible faster than $v$ itself, the particle is harder to stop.

Q1 (2 marks): $\dot v = -kv^2 \Rightarrow dv/v^2 = -k\,dt$; integrate $-1/v = -kt + C$ [1]. $v(0) = v_0$: $C = -1/v_0$; rearrange to $v(t) = v_0/(1 + k v_0 t)$ [1].

Q2 (3 marks): $m\dot v = mg - k v^2$; with $v_T^2 = mg/k$, $\dot v = (g/v_T^2)(v_T^2 - v^2)$ [1]. Partial fractions and integration give $\ln|(v_T + v)/(v_T - v)| = 2gt/v_T + C$; $v(0)=0$ gives $C = 0$ [1]. Solving: $v(t) = v_T \tanh(gt/v_T)$ [1].

Q3 (3 marks): (a) Upward positive, ascent: $v\frac{dv}{dy} = -g - kv^2$ [1]. (b) Separate $\frac{v\,dv}{g + kv^2} = -dy$; integrate from $(0,u)$ to $(H,0)$: $H = \frac{1}{2k}\ln\!\left(1 + \frac{ku^2}{g}\right)$ [1]. (c) On descent, motion is downward so $|v|$ is now negative under the upward-positive convention; the drag is $+kv^2$ (acting opposite to motion), so the equation becomes $v\frac{dv}{dy} = -g + kv^2$, a sign change [1].

Consolidate and move on

Sit the module quiz, then close the lesson off.

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 09 Lesson 11 →

Module overview · Extension 2