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.
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?
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.
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: $v_T = mg/k$
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
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
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
Integrating from $0$ to $t$ with $v_x(0) = u$:
A second integration (with $x(0) = 0$) gives the horizontal displacement:
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).
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 $t$?
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
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):
Applying $v_y(0) = w_0$ (initial vertical velocity) and inverting:
Terminal velocity. As $t \to \infty$, the exponential dies and $v_y \to -mg/k$. The downward terminal speed is
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 $v_T = mg/k$.
Worked examples · 3 in a row, reveal as you go
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.
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.
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.
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 = $ .
Misconceptions to fix · the 3 traps that cost marks
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.
Activities · practice with the ideas
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.
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.
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 $\frac{1}{k}\ln(1 + ku/g)$.
For the previous projectile, find the maximum height reached above the launch point.
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?
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.
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 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)
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)
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)
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].
Five timed questions on linear-drag projectile motion, terminal velocity, and component ODEs. 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.
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