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hscscience Ext 2 · Y12
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Module 16 · L09 of 16 ~45 min ⚡ +90 XP available

Projectile Motion with Air Resistance (Proportional to v)

A projectile in air does not follow a perfect parabola, drag bleeds energy and bends the path. When resistance is proportional to speed (a good model at low velocities), the horizontal and vertical components decouple and each becomes a first-order linear ODE. This lesson shows how to set up the equations, solve them, and read terminal velocity straight from the structure.

Today's hook, A skydiver in free fall does not accelerate forever, they reach a steady terminal velocity. Write Newton's second law for the descent, assuming air resistance proportional to speed: $m\dot v = mg - kv$. Before reading on, predict the terminal velocity $v_T$ in terms of $m, g, k$ by setting $\dot v = 0$. Compare your answer after card 05.
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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).

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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 $kv$ (opposing motion). Write Newton's second law as a differential equation in $v$. Before checking solve it for $v(t)$ given $v(0) = v_0$. What kind of decay is this?

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The two moves for projectile + linear drag

+5 XP to read When drag is proportional to velocity, every projectile problem reduces to two habits: resolve into horizontal...

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The two moves for projectile + linear drag
+5 XP to read

When drag is proportional to velocity, every projectile problem reduces to two habits: resolve into horizontal and vertical components (they decouple completely because the drag terms $-kv_x$ and $-kv_y$ are linear), then solve each first-order linear ODE using the integrating factor or separation of variables. Mixing the components or forgetting gravity's sign is the most common error.

The decouple-and-solve reading: (1) set $x$ horizontal, $y$ vertical (upward positive), (2) apply Newton II to each component, gravity acts only on $y$, drag acts on both, (3) solve each first-order ODE separately and integrate to get position.

Horizontal: $\dot v_x = -k v_x$  ·  Vertical: $\dot v_y = -g - k v_y$  (rising) or $-g + k v_y$ wording depends on convention.

Migrated 'Decouple components flow' as a 4-step sequence.
$m\ddot x = -k\dot x \;,\quad m\ddot y = -mg - k\dot y$
Drag opposes velocity
The drag term is $-kv$ (with $v$ a signed component). On the way up, $v_y > 0$ so drag points down; on the way down, $v_y < 0$ so $-kv_y > 0$, drag flips to point up. One equation handles both phases.
Components decouple
Because drag is linear in $v$, the horizontal ODE involves only $v_x$ and the vertical only $v_y$. You can solve each in isolation, no system to invert.
Terminal velocity = equilibrium
In the falling phase, set $\dot v_y = 0$. The equation $-g + k v_T / m = 0$ gives $v_T = mg/k$. This is the long-term limit, approached exponentially.
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What you'll master
Know

Key facts

  • Linear drag model: force $= -kv$, with $k > 0$ a positive constant
  • Horizontal equation: $m\ddot x = -k\dot x$, solution $v_x = v_0 e^{-kt/m}$
  • Vertical equation: $m\ddot y = -mg - k\dot y$ (taking upward as positive)
  • Terminal velocity: $v_T = mg/k$
Understand

Concepts

  • Why horizontal and vertical components decouple under linear drag
  • Why the projectile path is not symmetric about the apex
  • Why terminal velocity exists and is approached but never reached
Can do

Skills

  • Set up Newton II equations for both components with correct signs
  • Solve first-order linear ODEs by separation of variables
  • Integrate $v(t)$ to obtain $x(t)$ and $y(t)$ with initial conditions
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Key terms
Linear dragResistive force of magnitude $kv$, proportional to the first power of speed. A good model for slow motion in viscous fluids (Stokes drag).
Resistance constant $k$A positive constant carrying the units of mass per time. Determined by the geometry of the projectile and the viscosity of the medium.
Terminal velocity $v_T$The constant speed at which gravity is exactly balanced by drag. For linear drag, $v_T = mg/k$. Approached exponentially as $t \to \infty$.
Separable ODEA first-order equation that can be written $\frac{dv}{dt} = f(v)g(t)$, then solved by $\int \frac{dv}{f(v)} = \int g(t)\,dt$.
Horizontal rangeUnder linear drag, $x(t) = \frac{mv_0\cos\theta}{k}(1 - e^{-kt/m})$, bounded above by $\frac{m v_0 \cos\theta}{k}$ no matter how long the flight.
Asymmetry of trajectoryThe descent is steeper than the ascent because the projectile loses horizontal speed throughout, so the apex sits closer to the landing point than the launch.
MEX-M1NESA outcome (Applications of Calculus to Mechanics): solves problems involving resisted motion of a projectile, including resistance proportional to velocity.
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Horizontal component, pure exponential decay

Project a particle of mass $m$ with initial horizontal velocity $u = v_0 \cos\theta$.

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Horizontal component, pure exponential decay
core concept

Project a particle of mass $m$ with initial horizontal velocity $u = v_0 \cos\theta$. The only horizontal force is drag $-k v_x$ (gravity is vertical). Newton II gives

$$m \frac{dv_x}{dt} = -k v_x \quad\Longrightarrow\quad \frac{dv_x}{v_x} = -\frac{k}{m}\,dt.$$

Integrating from $0$ to $t$ with $v_x(0) = u$:

$$v_x(t) = u\, e^{-kt/m}.$$

A second integration (with $x(0) = 0$) gives the horizontal displacement:

$$x(t) = \frac{m u}{k}\bigl(1 - e^{-kt/m}\bigr).$$

Worked through the hook: for the skydiver falling vertically, $v_x = 0$, so the horizontal equation gives $x = 0$, they fall straight down. The interesting equation is the vertical one (next card).

Connecting to limits. As $t \to \infty$, $v_x \to 0$ and $x \to mu/k$. The horizontal range is bounded: the particle cannot travel further than $mu/k$ horizontally, no matter how long it stays aloft.

Horizontal ODE: $m\dot v_x = -kv_x$, separable, exponential decay · $v_x(t) = u e^{-kt/m}$ and $x(t) = (mu/k)(1 - e^{-kt/m})$ · Horizontal range bounded above by $mu/k$ · Gravity does not appear in the horizontal equation, only drag

Pause, copy the horizontal ODE $m\dot v_x = -kv_x$, the solution $v_x = u e^{-kt/m}$, the position $x = (mu/k)(1-e^{-kt/m})$, and the bounded-range limit $mu/k$ into your book.

Quick check: A particle is projected horizontally with initial velocity $u$ and experiences air resistance of magnitude $kv$ per unit mass (so $\dot v = -kv$). What is the velocity at time $t$?

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Vertical component & terminal velocity

We just saw the horizontal component under linear drag: $v_x = u e^{-kt/m}$ (exponential decay), $x = (mu/k)(1-e^{-kt/m})$, with horizontal range bounded above by $mu/k$, gravity plays no role horizontally.

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Vertical component & terminal velocity
core concept

We just saw the horizontal component under linear drag: $v_x = u e^{-kt/m}$ (exponential decay), $x = (mu/k)(1-e^{-kt/m})$, with horizontal range bounded above by $mu/k$, gravity plays no role horizontally. That raises a question: what happens to the vertical component, and what is terminal velocity? This card answers it → the vertical ODE adds $-mg$ to the drag; solution gives $v_y \to -mg/k$ (terminal, downward) as $t \to \infty$.

Take upward as positive. The vertical forces are gravity $-mg$ and drag $-k v_y$ (drag always opposes the signed velocity). Newton II gives

$$m \frac{dv_y}{dt} = -mg - k v_y.$$

This is a first-order linear ODE. Rearrange and separate (assume the falling phase, $v_y < 0$, and let $w = -v_y > 0$ for clarity, or solve directly):

$$\frac{dv_y}{-g - (k/m)v_y} = dt \;\Longrightarrow\; -\frac{m}{k}\ln\!\bigl|mg + k v_y\bigr| = t + C.$$

Applying $v_y(0) = w_0$ (initial vertical velocity) and inverting:

$$v_y(t) = -\frac{mg}{k} + \left(w_0 + \frac{mg}{k}\right) e^{-kt/m}.$$

Terminal velocity. As $t \to \infty$, the exponential dies and $v_y \to -mg/k$. The downward terminal speed is

$$v_T = \frac{mg}{k}.$$
Equilibrium intuition. Terminal velocity is the value of $v_y$ for which $\dot v_y = 0$ in the falling phase: gravity pulling down ($mg$) equals drag pushing up ($kv_T$). Setting $mg = k v_T$ recovers $v_T = mg/k$ instantly, no integration needed.

Vertical ODE (upward positive): $m\dot v_y = -mg - kv_y$ · Solution: $v_y(t) = -\frac{mg}{k} + \left(w_0 + \frac{mg}{k}\right)e^{-kt/m}$ · Terminal velocity: $v_T = mg/k$ (downward) · Set $\dot v_y = 0$ in the falling phase to get $v_T$ without solving the ODE

Pause, copy the vertical ODE $m\dot v_y = -mg-kv_y$, the solution form, the terminal velocity $v_T = mg/k$, and the shortcut $\dot v_y = 0 \Rightarrow v_T$ into your book.

Did you get this? True or false: for a body falling from rest under gravity with air resistance proportional to speed ($m\dot v = mg - kv$, downward positive), the terminal velocity is $v_T = mg/k$.

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

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

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

PROBLEM 1 · HORIZONTAL PROJECTILE

A particle is projected horizontally with initial speed $20\text{ m/s}$. Air resistance per unit mass is $0.5 v$. Find $v(t)$ and the limiting horizontal displacement.

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Newton II per unit mass: $\dot v = -0.5 v$. Separate: $dv/v = -0.5\,dt$, integrate: $\ln v = -0.5 t + C$.
No gravity acts horizontally. Separation of variables is the cleanest route for $\dot v = -kv$.
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PROBLEM 2 · BODY FALLING FROM REST

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

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Newton II: $m\dot v = mg - kv$. Rearrange: $\dot v = g - (k/m)v$. Equilibrium at $\dot v = 0$ gives $v_T = mg/k$.
Reading $v_T$ off the equation before solving keeps the algebra honest. Any solution must approach $v_T$ as $t \to \infty$.
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PROBLEM 3 · TIME TO HALF TERMINAL VELOCITY

Using $v(t) = \frac{mg}{k}(1 - e^{-kt/m})$ for a body falling from rest, find the time taken to reach half of terminal velocity.

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Half of terminal velocity: $\tfrac12 v_T = \tfrac{1}{2} \cdot \tfrac{mg}{k}$. Set $v(t) = \tfrac12 v_T$.
Define the target first; then equate. Substituting numerical fractions of $v_T$ keeps the algebra dimensionless.
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Fill the gap: A body falls from rest with air resistance $kv$ per unit mass. The terminal velocity is $v_T = $ , and the time taken to reach half this speed is $t = $ .

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Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Mixing up the direction convention
If upward is positive, gravity is $-mg$ and the drag is $-kv_y$ (always opposing signed velocity). If downward is positive, the same body has gravity $+mg$ and drag $-kv$. Switching mid-question flips signs, fix one convention and stick to it.
Trap 02
Forgetting that horizontal motion has no gravity term
The horizontal equation is $m\dot v_x = -k v_x$, gravity is purely vertical and does not appear. Including $-mg$ in the horizontal ODE is a common slip when working under time pressure.
Trap 03
Claiming the body reaches terminal velocity
Strictly, $v_y$ approaches $v_T$ asymptotically, it never equals it. Solutions like "after 10 seconds the body has reached terminal velocity" should say "is within $\varepsilon$ of terminal velocity" or "the relative error is less than 1%".

Did you get this? True or false: under linear air resistance, the horizontal and vertical velocity components satisfy a coupled pair of ODEs that must be solved simultaneously.

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Revisit your thinking

Activities · practice with the ideas

Activities · practice with the ideas

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A particle is projected horizontally with speed $30\text{ m/s}$. Resistance per unit mass is $0.2 v$. Find $v(t)$ and the limiting horizontal distance.

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A body of mass $2$ kg falls from rest. Air resistance is $4v$ N. Take $g = 10\text{ m/s}^2$. Find the terminal velocity and the time to reach $90\%$ of it.

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A particle is projected vertically upward with speed $u$ and experiences air resistance $kv$ per unit mass. Show that the time taken to reach the highest point is $\frac{1}{k}\ln(1 + ku/g)$.

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For the previous projectile, find the maximum height reached above the launch point.

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Explain qualitatively why a projectile launched at $45^\circ$ no longer achieves maximum range when air resistance is present.

Odd one out: Three of these are valid expressions of the terminal velocity for a body falling under gravity with air resistance proportional to speed. Which one is NOT?

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Revisit your thinking

Earlier you wrote Newton II for a skydiver with drag $kv$ and predicted the terminal velocity by setting $\dot v = 0$.

The answer is $v_T = mg/k$, and you can read this directly off the equation $m\dot v = mg - kv$ without solving the ODE. The full solution $v(t) = (mg/k)(1 - e^{-kt/m})$ confirms the limit and shows the approach is exponential with time constant $\tau = m/k$. The "half-time" to terminal velocity is $\tau \ln 2$, the same structure as radioactive decay. Recognising terminal velocity as the equilibrium of a first-order linear ODE is the single most reusable idea in Module 16.

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Show what you have learned

Multiple choice, then short answer under exam conditions.

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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.

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Short answer
ApplyBand 32 marks

Q1. A particle is projected horizontally with initial speed $u$. Resistance per unit mass is $kv$. Show that $v(t) = u e^{-kt}$ and find the limiting horizontal distance. (2 marks)

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ApplyBand 43 marks

Q2. A body of mass $m$ falls from rest under gravity with air resistance $kv$ (downward positive). Derive $v(t) = (mg/k)(1 - e^{-kt/m})$ and state the terminal velocity. (3 marks)

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AnalyseBand 53 marks

Q3. A particle of mass $m$ is projected vertically upward with speed $u$ and air resistance is $kv$ per unit mass. (a) Write the equation of motion for the ascent. (b) Find the time to reach the highest point. (c) Explain why the descent time exceeds the ascent time. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $\dot v = -0.2 v \Rightarrow v(t) = 30 e^{-0.2 t}$. Then $x(t) = 150(1 - e^{-0.2t})$ and $x \to 150\text{ m}$.

2. $v_T = mg/k = (2)(10)/4 = 5\text{ m/s}$. $v(t) = 5(1 - e^{-2t})$; set $1 - e^{-2t} = 0.9 \Rightarrow t = \tfrac12 \ln 10 \approx 1.15\text{ s}$.

3. Upward positive: $\dot v = -g - kv$. Separate: $\int \frac{dv}{g + kv} = -\int dt \Rightarrow \frac{1}{k}\ln(g + kv) = -t + C$. $v(0)=u$ gives $C = \frac{1}{k}\ln(g+ku)$. Setting $v(T)=0$ yields $T = \frac{1}{k}\ln\!\left(1 + \frac{ku}{g}\right)$.

4. Use $v\frac{dv}{dy} = -g - kv$. Separate $\frac{v\,dv}{g + kv} = -dy$, split $\frac{v}{g+kv} = \frac{1}{k} - \frac{g}{k(g+kv)}$, integrate from $(0,u)$ to $(H,0)$: $H = \frac{u}{k} - \frac{g}{k^2}\ln\!\left(1 + \frac{ku}{g}\right)$.

5. Drag bleeds horizontal speed continuously, so range is gained more efficiently by launching with a larger horizontal component. The optimal angle drops below $45^\circ$.

Q1 (2 marks): Separate $dv/v = -k\,dt$; integrate to $\ln v = -kt + \ln u$, so $v = u e^{-kt}$ [1]. Then $x = \int_0^t u e^{-ks}ds = (u/k)(1 - e^{-kt}) \to u/k$ [1].

Q2 (3 marks): $m\dot v = mg - kv$ [1]. Separate $dv/(mg/k - v) = (k/m)dt$ (or equivalent), integrate, $v(0) = 0$ gives $v(t) = (mg/k)(1 - e^{-kt/m})$ [1]. As $t \to \infty$, $v \to mg/k$, the terminal velocity [1].

Q3 (3 marks): (a) $\dot v = -g - kv$ (upward positive) [1]. (b) Separate, integrate, $v(0)=u$, set $v(T)=0$: $T = \frac{1}{k}\ln(1 + ku/g)$ [1]. (c) On the way up, gravity and drag both decelerate; on the way down, gravity accelerates while drag decelerates, so $|a_{\text{down}}| < g$. Same distance, smaller acceleration $\Rightarrow$ longer time [1].