Resisted Motion — Vertical Falling
When a body falls through air or fluid, gravity pulls it down while resistance pushes back. The faster it falls, the harder resistance fights — until the two forces balance and acceleration vanishes. That balance point is the terminal velocity. In this lesson you'll set up the equation of motion $\dot v = g - kv$ (and the quadratic form $\dot v = g - kv^2$), solve it by separation of variables, and answer the canonical HSC question: how long does it take to reach a given fraction of terminal velocity?
A body of mass $m$ falls under gravity with linear resistance proportional to its speed. Take down as positive. Before checking — write the equation of motion using Newton's second law, then solve $\dot v = 0$ to find the terminal speed in terms of $g$ and the resistance constant.
Every vertical resisted-motion problem rewards two habits: choose a sign convention (down positive is standard for a falling object), then match the calculus form to the question — use $\dfrac{dv}{dt}$ when time is wanted, $v\dfrac{dv}{dx}$ when distance is wanted. Picking the wrong form turns a one-line separation into a tangled mess.
The convention-equation-form reading: (1) fix downward as positive, (2) write $m\dot v = mg - R(v)$ where $R(v)$ is the resistance, (3) pick $\dfrac{dv}{dt}$ (time question) or $v\dfrac{dv}{dx}$ (displacement question).
Linear: $\dot v = g - kv$ · Quadratic: $\dot v = g - kv^2$ · Terminal: $\dot v = 0$
Key facts
- Equation of motion for vertical fall with linear resistance: $\dot v = g - kv$
- Equation of motion with quadratic resistance: $\dot v = g - kv^2$
- Terminal velocity $v_T$ found by setting $\dot v = 0$
- $v_T = g/k$ (linear) and $v_T = \sqrt{g/k}$ (quadratic)
Concepts
- Why resistance opposes velocity and changes sign if the body reverses
- Why $v \to v_T$ asymptotically but is never reached in finite time
- Why $v\dfrac{dv}{dx}$ is the right form for displacement questions
Skills
- Set up Newton's second law for a falling body with resistance
- Solve $\dot v = g - kv$ by separation of variables, applying $v(0)=0$
- Find the time to reach a given fraction $\alpha v_T$ of terminal velocity
- Use $v\dfrac{dv}{dx}$ to find distance fallen to reach a target speed
Take downward as positive. Two forces act on a body of mass $m$ falling from rest: gravity $mg$ (down, +) and resistance $R(v)$ (up, $-$, because the body moves down). Newton's second law gives
For the linear case $R(v) = mkv$ (so the constant $k$ has units $\text{s}^{-1}$):
Terminal velocity. Set $\dot v = 0$: $g - kv_T = 0$, so $v_T = g/k$.
Separation of variables (starting from rest, $v(0)=0$):
- $\dfrac{dv}{g - kv} = dt$.
- Integrate: $-\dfrac{1}{k}\ln|g - kv| = t + C$.
- At $t = 0$, $v = 0$: $C = -\dfrac{1}{k}\ln g$.
- Combine: $\ln\!\left(\dfrac{g}{g - kv}\right) = kt$, so $g - kv = g\,e^{-kt}$.
- Solve: $v(t) = \dfrac{g}{k}\bigl(1 - e^{-kt}\bigr) = v_T\bigl(1 - e^{-kt}\bigr)$.
Worked through the hook. With the diver of mass $m$ and resistance $mkv$, the terminal velocity is $v_T = g/k$. Because $v(t) = v_T(1 - e^{-kt}) < v_T$ for every finite $t$, the diver never reaches the terminal velocity — the curve approaches it asymptotically. HSC questions therefore ask for the time to reach a fraction such as $0.9\,v_T$.
Down positive. $m\dot v = mg - R(v)$. · Linear case: $\dot v = g - kv$; $v_T = g/k$; $v(t) = v_T(1 - e^{-kt})$. · Set $\dot v = 0$ to find $v_T$. · $v \to v_T$ as $t \to \infty$ but never equals it in finite time.
Pause — copy the linear-falling ODE $m\dot v = mg-kv$, $v_T = g/k$, $v(t) = v_T(1-e^{-kt})$, and the rule to set $\dot v = 0$ for $v_T$ into your book.
Quick check: A body falls from rest under gravity with linear resistance, satisfying $\dot v = g - kv$ with $g, k > 0$. What is the terminal velocity?
We just saw the linear-drag falling ODE $m\dot v = mg-kv$, giving $v_T = g/k$ (linear ratio) and $v(t) = v_T(1-e^{-kt})$ — terminal approached but never reached in finite time. That raises a question: for quadratic resistance $R = kv^2$, how do we find the displacement as a function of velocity? This card answers it → use $v\,dv/dx = g-kv^2$, factor as $k(v_T-v)(v_T+v)$, integrate by partial fractions to get $x = (1/2k)\ln(v_T^2/(v_T^2-v^2))$.
At higher speeds resistance is better modelled by $R(v) = mkv^2$, giving
Setting $\dot v = 0$: $v_T^2 = g/k$, so $v_T = \sqrt{g/k}$. The factored form makes integration tractable:
For distance fallen, swap to the chain-rule form $\dot v = v\dfrac{dv}{dx}$:
The left side integrates by inspection: let $u = g - kv^2$, then $du = -2kv\,dv$, so $\displaystyle\int \dfrac{v\,dv}{g-kv^2} = -\dfrac{1}{2k}\ln|g - kv^2|$. Cleanly applying $v(0) = 0$ gives
Quadratic: $\dot v = g - kv^2$; terminal $v_T = \sqrt{g/k}$. · Use $v\dfrac{dv}{dx}$ when distance is wanted, $\dfrac{dv}{dt}$ when time is wanted. · $g - kv^2 = k(v_T - v)(v_T + v)$ — factored form helps partial fractions. · $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$ (distance fallen, quadratic case, from rest).
Pause — copy the quadratic-falling ODE $m\dot v = mg-kv^2$, terminal $v_T = \sqrt{g/k}$, the factored form $k(v_T-v)(v_T+v)$, the partial-fraction split, and the displacement formula $x = (1/2k)\ln(v_T^2/(v_T^2-v^2))$ into your book.
Did you get this? True or false: a body falling from rest with linear resistance $\dot v = g - kv$ reaches the terminal velocity $v_T = g/k$ in finite time.
Worked examples · 3 in a row, reveal as you go
A particle falls from rest with linear resistance $\dot v = g - kv$. Find the time taken to reach $90\%$ of the terminal velocity, expressing the answer in terms of $k$.
A particle falls from rest with $\dot v = g - kv^2$. Find the distance fallen by the time the speed reaches $\tfrac{1}{2} v_T$, where $v_T = \sqrt{g/k}$.
A body of mass $m$ is projected downward at speed $u$ where $u < v_T = g/k$, with linear resistance $\dot v = g - kv$. Show that $v(t) = v_T - (v_T - u)e^{-kt}$, and find the value of $u$ for which the body falls with constant velocity from the outset.
Fill the gap: For a body falling from rest with $\dot v = g - kv$, the velocity is $v(t) = v_T(1 - e^{})$, where the terminal velocity is $v_T = $.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $\dot v = g - kv^2$ the terminal velocity is $v_T = g/k$.
Activities · practice with the ideas
A particle of mass $m$ falls from rest with resistance $mkv$. Write the equation of motion (down positive) and find the terminal velocity in terms of $g$ and $k$.
Starting from $v(t) = v_T(1 - e^{-kt})$, find the time taken to reach $\tfrac{1}{2} v_T$ in terms of $k$.
For $\dot v = g - kv^2$ falling from rest, derive an expression for the distance fallen $x$ in terms of $v$, $g$ and $k$.
A body is projected downwards at $u = 2 v_T$ in a linear-resistance medium with $\dot v = g - kv$. Will the body speed up or slow down? Write the equation for $v(t)$.
Show that the distance fallen as $v \to v_T$ for the quadratic-resistance case becomes infinite, and explain physically.
Odd one out: Three of these are correct facts about a body falling from rest with $\dot v = g - kv$. Which is NOT?
Earlier you set up Newton's second law for a falling body with resistance, and you predicted whether the diver could actually reach $v_T$.
The equation $\dot v = g - kv$ gives $v(t) = v_T(1 - e^{-kt})$, so the diver's speed climbs exponentially toward $v_T = g/k$ but never quite gets there: $v < v_T$ for every finite $t$. The HSC question is therefore framed as "time to reach $0.9\,v_T$" or similar — never "time to reach $v_T$". Recognising this distinction is the single most useful sanity check 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 of mass $m$ falls from rest in a medium where the resistance is $mkv$ ($k > 0$). Write the equation of motion using $\dfrac{dv}{dt}$ and state the terminal velocity. (2 marks)
Q2. For a body falling from rest with $\dot v = g - kv$, show that $v(t) = \dfrac{g}{k}(1 - e^{-kt})$ and hence find the time taken to reach $80\%$ of the terminal velocity. (3 marks)
Q3. A particle falls from rest under gravity in a medium where the resistance per unit mass is $kv^2$, so $\dot v = g - kv^2$. (a) Find the terminal velocity $v_T$. (b) Show that the distance fallen as a function of speed is $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Down positive: $m\dot v = mg - mkv \Rightarrow \dot v = g - kv$. Setting $\dot v = 0$: terminal velocity $v_T = g/k$.
2. $v = \tfrac{1}{2}v_T$: $\tfrac{1}{2} = 1 - e^{-kt} \Rightarrow e^{-kt} = \tfrac{1}{2} \Rightarrow t = \dfrac{\ln 2}{k} \approx \dfrac{0.693}{k}$.
3. $v\dfrac{dv}{dx} = g - kv^2$. Separate, $u = g - kv^2$, $du = -2kv\,dv$: $x = -\dfrac{1}{2k}\ln(g - kv^2) + C$. From $v(0)=0,\;x=0$: $C = \dfrac{1}{2k}\ln g$. So $x = \dfrac{1}{2k}\ln\!\left(\dfrac{g}{g - kv^2}\right)$.
4. At $v = 2v_T = 2g/k$: $\dot v = g - k(2g/k) = -g < 0$, so the body slows. General solution with $v(0) = 2v_T$: $v(t) = v_T - (v_T - 2v_T)e^{-kt} = v_T + v_T e^{-kt} = v_T(1 + e^{-kt})$. As $t \to \infty$, $v \to v_T$ from above.
5. From Q3, $x = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$. As $v \to v_T^-$, $v_T^2 - v^2 \to 0^+$, so the logarithm $\to +\infty$ and $x \to \infty$. Physically the body never reaches $v_T$ in finite time, so it must fall an unbounded distance to approach it.
Q1 (2 marks): Equation $\dot v = g - kv$ [1]. Terminal velocity: $\dot v = 0 \Rightarrow v_T = g/k$ [1].
Q2 (3 marks): Separation: $\dfrac{dv}{g - kv} = dt$; integration: $-\tfrac{1}{k}\ln|g - kv| = t + C$; IC $v(0)=0 \Rightarrow C = -\tfrac{1}{k}\ln g$; rearrange to $v = \dfrac{g}{k}(1 - e^{-kt})$ [2 marks]. For $v = 0.8\,v_T$: $0.8 = 1 - e^{-kt} \Rightarrow e^{-kt} = 0.2 \Rightarrow t = \dfrac{\ln 5}{k}$ [1].
Q3 (3 marks): (a) $\dot v = 0 \Rightarrow kv_T^2 = g \Rightarrow v_T = \sqrt{g/k}$ [1]. (b) Use $\dot v = v\,dv/dx$: $\dfrac{v\,dv}{g - kv^2} = dx$. Sub $u = g - kv^2$: $x = -\dfrac{1}{2k}\ln(g - kv^2) + C$ [1]. From rest: $C = \dfrac{1}{2k}\ln g$; therefore $x = \dfrac{1}{2k}\ln\!\left(\dfrac{g}{g - kv^2}\right) = \dfrac{1}{2k}\ln\!\left(\dfrac{v_T^2}{v_T^2 - v^2}\right)$ [1].
Five timed questions on resisted vertical motion, terminal velocity, and fractional-time problems. 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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