Throw a ball straight up. Gravity pulls it down; air drag also pulls it down (because drag opposes the upward velocity). Both forces work against the motion, so the deceleration is larger than $g$. This lesson teaches you to set up $\dot v = -g - kv$ (or $-g - kv^2$), integrate to find the time of maximum height and use $v\,dv/dx$ to find the maximum height itself.
+5 XP warm-up A ball is thrown vertically upward with speed in a vacuum (no drag).
A ball is thrown vertically upward with speed $u$ in a vacuum (no drag). Recall: time to max height is $u/g$; max height is $u^2/(2g)$. Now imagine air drag is switched on. Will both of these numbers go up or down? Why?
+5 XP to read Every rising-with-resistance problem rewards two habits: choose "up positive" and write both downward forces...
Every rising-with-resistance problem rewards two habits: choose "up positive" and write both downward forces with minus signs, then pick $dv/dt$ for time questions or $v\,dv/dx$ for height questions. Both gravity and drag are negative while $v > 0$, this is the most-missed sign.
The sign-derivative-target reading: (1) "up" is positive, so $g$ and the drag term both carry minus signs while rising; (2) decide whether the target is time or height; (3) use $dv/dt$ to find time to apex, $v\,dv/dx$ to find height of apex.
Rising (linear drag): $\dfrac{dv}{dt} = -g - kv$ · Rising (quadratic): $\dfrac{dv}{dt} = -g - kv^2$ · Max height when $v = 0$.
A particle of mass $m$ is thrown vertically upward at speed $u$.
A particle of mass $m$ is thrown vertically upward at speed $u$. Take $x$ positive upward. Net downward force while rising is weight plus drag: $mg + mkv$. Newton's second law:
Separate variables: $\dfrac{dv}{g+kv} = -dt$. Integrate from $v = u$ to $v$:
For the height, switch to $v\,dv/dx = -(g+kv)$, separate as $\dfrac{v\,dv}{g+kv} = -dx$, and integrate from $v = u$, $x = 0$ to $v = 0$, $x = H$. The standard split $\dfrac{v}{g+kv} = \dfrac{1}{k} - \dfrac{g/k}{g+kv}$ gives:
Rising (linear drag): $\dot v = -(g + kv)$, both forces point down while $v > 0$ · $v(t) = \dfrac{(g+ku)e^{-kt} - g}{k}$ · Time to apex: $t^* = \dfrac{1}{k}\ln\!\left(1 + \dfrac{ku}{g}\right) < u/g$ · Apex height: $H = \dfrac{u}{k} - \dfrac{g}{k^2}\ln\!\left(1 + \dfrac{ku}{g}\right) < \dfrac{u^2}{2g}$
Pause, copy the linear-drag rising ODE $\dot v = -(g+kv)$, the time-to-apex $t^* = (1/k)\ln(1+ku/g)$, the apex height $H = u/k - (g/k^2)\ln(1+ku/g)$, and the apex-height inequality $H < u^2/(2g)$ into your book.
Quick check: A particle is projected vertically upward at $u$ m/s and decelerates under $\dot v = -g - kv$. The time $t^*$ to reach maximum height is:
We just saw rising under linear drag : decays to rest in finite time at height \dot v = -(g+kv^2)t^*H\int dv/(g+kv^2) =...
We just saw rising under linear drag $\dot v = -(g+kv)$: $v(t)$ decays to rest in finite time $t^* = (1/k)\ln(1+ku/g)$ at height $H < u^2/(2g)$, drag shortens range. That raises a question: how does the quadratic-drag case $\dot v = -(g+kv^2)$ differ, particularly the integral needed for $t^*$ and $H$? This card answers it → $\int dv/(g+kv^2) = (1/\sqrt{gk})\arctan(v\sqrt{k/g})$ gives $t^* = (1/\sqrt{gk})\arctan(u\sqrt{k/g})$ and $H = (1/2k)\ln(1+ku^2/g)$.
For larger projectiles, drag $\propto v^2$. The equation of motion while rising (up positive) becomes:
Separate: $\dfrac{dv}{g+kv^2} = -dt$. Factor out $g$ to put the integrand in $\arctan$ form. Let $\alpha = \sqrt{g/k}$; then:
Apex height: from $v = u$, $x = 0$ to $v = 0$, $x = H$:
Rising (quadratic drag): $\dot v = -(g + kv^2)$ · $\int \dfrac{dv}{g+kv^2} = \dfrac{1}{\sqrt{gk}}\arctan(v\sqrt{k/g})$ · Time to apex: $t^* = \dfrac{1}{\sqrt{gk}}\arctan\!\left(u\sqrt{\tfrac{k}{g}}\right)$ · Apex height: $H = \dfrac{1}{2k}\ln\!\left(1 + \dfrac{ku^2}{g}\right)$
Pause, copy the quadratic-drag integral $\int dv/(g+kv^2) = (1/\sqrt{gk})\arctan(v\sqrt{k/g})$, $t^* = (1/\sqrt{gk})\arctan(u\sqrt{k/g})$, and $H = (1/2k)\ln(1+ku^2/g)$ into your book.
Did you get this? True or false: in the model $\dot v = -g - kv$ (up positive), the time to reach maximum height is strictly less than $u/g$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
A ball of unit mass is thrown upward at $u = 20$ m/s. Air resistance gives $\dot v = -10 - \tfrac12 v$ (taking $g = 10$). Find the time to maximum height.
Continuing from Problem 1 ($u = 20$, $g = 10$, $k = \tfrac12$), use $v\,dv/dx$ to find the maximum height reached.
A projectile of unit mass is fired straight up at $u = 30$ m/s. Drag gives $\dot v = -g - \tfrac{1}{90} v^2$ with $g = 10$. Find the maximum height.
Fill the gap: For rising motion under quadratic drag, $\dot v = -g - kv^2$, the maximum height is $H = \dfrac{1}{p}\ln\!\left(1 + \dfrac{ku^2}{q}\right)$. Give $p =$ and $q =$ .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: while a ball rises under linear drag with up-positive convention, both gravity and air resistance contribute negative terms to $dv/dt$.
Activities · practice with the ideas
Activities · practice with the ideas
A particle is thrown up at $u = 15$ m/s under $\dot v = -10 - v$ (take $g = 10$, $k = 1$). Find the time to maximum height.
For the same setup, find the maximum height using $v\,dv/dx$.
Derive the general formula $H = \dfrac{u}{k} - \dfrac{g}{k^2}\ln\!\left(1 + \dfrac{ku}{g}\right)$ from $v\,dv/dx = -(g+kv)$.
Quadratic drag: a particle of unit mass has $\dot v = -10 - \tfrac{v^2}{40}$ and $u = 20$. Find the maximum height.
For quadratic drag, derive $t^* = \dfrac{1}{\sqrt{gk}}\arctan\!\left(u\sqrt{k/g}\right)$ as the time to reach maximum height, starting from $\dot v = -(g+kv^2)$.
Odd one out: Three of these statements about a ball rising under linear drag (up positive) are correct. Which one is NOT?
Earlier you predicted whether drag makes the time-to-apex and the apex height smaller or larger than the vacuum values $u/g$ and $u^2/(2g)$.
Drag strictly reduces both: $t^* = \dfrac{1}{k}\ln(1 + ku/g) < u/g$ and $H = \dfrac{u}{k} - \dfrac{g}{k^2}\ln(1 + ku/g) < \dfrac{u^2}{2g}$. The physical reason: while rising, gravity and drag both point downward, so the deceleration is greater than $g$ alone. After the apex the signs change, the rising equation is only valid for $v > 0$, a subtle but examinable point.
Multiple choice, then short answer under exam conditions.
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 unit mass is projected vertically upward at $u = 10$ m/s. Resistance gives $\dot v = -g - v$ where $g = 10$. Show that the time to reach maximum height is $t^* = \ln 2$ seconds. (2 marks)
Q2. For the same particle, use $v\,\dfrac{dv}{dx}$ to find the maximum height reached. Compare with the vacuum value $u^2/(2g)$. (3 marks)
Q3. A projectile of unit mass moves vertically upward with equation $\dot v = -g - kv^2$, initial speed $u$. (a) Show that the time to reach maximum height is $\dfrac{1}{\sqrt{gk}}\arctan\!\left(u\sqrt{k/g}\right)$. (b) Hence find the maximum height in terms of $u$, $g$, $k$. (3 marks)
Activity answers:
1. $\dfrac{dv}{10+v} = -dt \Rightarrow \ln(10+v) = -t + \ln 25$. At apex $v = 0$: $\ln 10 = -t^* + \ln 25 \Rightarrow t^* = \ln 2.5 \approx 0.916$ s.
2. $\dfrac{v}{10+v} = 1 - \dfrac{10}{10+v}$. From $v=15$, $x=0$ to $v=0$, $x=H$: $H = 15 - 10\ln(25/10) = 15 - 10\ln 2.5 \approx 5.84$ m.
3. $\dfrac{v}{g+kv} = \dfrac{1}{k} - \dfrac{g/k}{g+kv}$. Integrating from $u$ to $0$ and $0$ to $H$ gives $H = \dfrac{u}{k} - \dfrac{g}{k^2}\ln\!\left(1+\dfrac{ku}{g}\right)$.
4. $v\,dv/dx = -(10 + v^2/40)$. After separation $\dfrac{1}{2}\ln(400+v^2) = -x/40 + C$, with $C = \tfrac{1}{2}\ln 800$. Apex: $H = 20\ln 2 \approx 13.86$ m.
5. $\dfrac{dv}{g+kv^2} = -dt$. $\int = \dfrac{1}{\sqrt{gk}}\arctan(v\sqrt{k/g})$. From $u$ to $0$: $t^* = \dfrac{1}{\sqrt{gk}}\arctan(u\sqrt{k/g})$.
Q1 (2 marks): $\dfrac{dv}{10+v} = -dt$, integrate: $\ln(10+v) = -t + \ln 20$ [1]. Apex $v = 0$: $\ln 10 = -t^* + \ln 20 \Rightarrow t^* = \ln 2$ s [1].
Q2 (3 marks): Use $v\,dv/dx = -(10+v)$, split $\dfrac{v}{10+v} = 1 - \dfrac{10}{10+v}$ [1]; integrating from $v=10$, $x=0$ to $v=0$, $x=H$: $H = 10 - 10\ln 2 \approx 3.07$ m [1]. Vacuum: $u^2/(2g) = 5$ m, so drag reduces apex by $\approx 1.93$ m [1].
Q3 (3 marks): (a) $\dfrac{dv}{g+kv^2} = -dt$, integrate using $\int \dfrac{dv}{g+kv^2} = \dfrac{1}{\sqrt{gk}}\arctan\!\left(v\sqrt{k/g}\right)$, evaluated from $u$ to $0$ gives $t^* = \dfrac{1}{\sqrt{gk}}\arctan\!\left(u\sqrt{k/g}\right)$ [1]. (b) Use $v\,dv/dx = -(g+kv^2)$; integrate $\dfrac{v\,dv}{g+kv^2}$ as $\dfrac{1}{2k}\ln(g+kv^2)$ [1]; evaluate from $v=u$, $x=0$ to $v=0$, $x=H$: $H = \dfrac{1}{2k}\ln\!\left(1 + \dfrac{ku^2}{g}\right)$ [1].
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