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.
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?
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: $\dot v = -g - (k/m)v^2$.
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: $v_T = \sqrt{mg/k}$
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
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
A particle of mass $m$ moves horizontally with initial speed $v_0 > 0$. The only force is drag $-k v^2$. Newton II gives
Integrating from $0$ to $t$ with $v(0) = v_0$:
A second integration (with $x(0) = 0$) gives
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 $mu/k$.
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 $v(t)$?
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
Setting $\dot v = 0$ gives the terminal velocity at once: $mg = k v_T^2$, so
To solve the ODE, rewrite using $v_T$ as $\dot v = (g/v_T^2)(v_T^2 - v^2)$ and separate:
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
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
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 $v = 5\text{ m/s}$.
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.
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.
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
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.
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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)
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)
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)
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].
Five timed questions on quadratic-drag projectile motion, $\tanh$ solutions, and terminal velocity. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick mechanics questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.