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Drag $cv^2$: coasting as $v_0/(1 + t/\tau)$, falling as $v_{\text{ter}}\tanh(gt/v_{\text{ter}})$ with $v_{\text{ter}} = \sqrt{mg/c}$, and why two-dimensional motion must be integrated numerically.
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By the end of this lesson you will be able to solve one-dimensional motion with quadratic drag, find terminal speeds and the time to approach them, and explain why a projectile with quadratic drag has no formula for its path.
The last lesson gave the drag on a sphere as $bv + cv^2$ with $c = \gamma D^2$, and the ratio $1.6 \times 10^3\,Dv$ that decides which term matters. You solved linear drag with exponentials and separated its $x$ and $y$ components. You can separate variables, use partial fractions, and you have met the hyperbolic functions $\sinh$, $\cosh$ and $\tanh$. This lesson solves the quadratic case, which covers nearly every object a person can see moving through air.
| Term | What it means |
|---|---|
| Quadratic drag | $f = cv^2$, opposite to the velocity, with $c = \gamma D^2$ for a sphere. |
| Drag equation | $f = \tfrac{1}{2}C_d\rho Av^2$, the engineer's form, with drag coefficient $C_d$ and frontal area $A$. |
| Terminal speed | $v_{\text{ter}} = \sqrt{mg/c}$ for quadratic drag. |
| Characteristic time | $\tau = m/cv_0$ for horizontal coasting: the time to lose half the speed. |
| Hyperbolic tangent | $\tanh x = (e^x - e^{-x})/(e^x + e^{-x})$, rising from $0$ toward $1$. |
| Coupled equations | Equations in which each component's rate depends on the other components. |
For a baseball, a car, a skydiver or a raindrop, the quadratic term dominates and the drag has magnitude $cv^2$. In one dimension the second law is still separable. Horizontally, $m\dot{v} = -cv^2$ integrates to
$$v(t) = \frac{v_0}{1 + t/\tau}, \qquad x(t) = v_0\tau\ln\left(1 + \frac{t}{\tau}\right), \qquad \tau = \frac{m}{cv_0}.$$
Vertically from rest, $m\dot{v} = mg - cv^2$ has terminal speed $v_{\text{ter}} = \sqrt{mg/c}$, and integrating with partial fractions gives
$$v(t) = v_{\text{ter}}\tanh\left(\frac{gt}{v_{\text{ter}}}\right), \qquad y(t) = \frac{v_{\text{ter}}^2}{g}\ln\cosh\left(\frac{gt}{v_{\text{ter}}}\right).$$
In two dimensions the drag is $-cv\vec{v}$, and its $x$ component $-c\sqrt{v_x^2 + v_y^2}\,v_x$ depends on $v_y$. The equations are coupled, and a projectile with quadratic drag has no formula for its path: it has to be computed numerically.
Another way: picture
Think of drag as the push of the air the body shoves aside every second. Double the speed and the body meets twice as much air per second, and gives each parcel twice the speed: four times the push. That steep growth caps a falling body at a terminal speed that depends only weakly on its mass, and it brakes a fast body hard at first and then ever more gently as it slows.
Another way: steps
Separating $m\,dv/dt = -cv^2$ gives $dv/v^2 = -(c/m)\,dt$, and integrating from $v_0$ at $t = 0$ gives $1/v - 1/v_0 = ct/m$. The reciprocal speed grows linearly in time. The speed halves at $t = \tau = m/cv_0$, falls to a third at $2\tau$ and to a quarter at $3\tau$: each halving takes longer than the one before, because the brake weakens as the square of the speed.
Integrating once more gives $x = v_0\tau\ln(1 + t/\tau)$, which grows without limit. That is the opposite of linear drag, where the body coasted a finite distance $v_0\tau$. Of course a real bicycle or a puck does stop, because rolling resistance or friction, roughly constant, takes over at low speed; quadratic drag alone never quite brings anything to rest in a finite distance.
Take down as positive and write the equation as $\dot{v} = g(1 - v^2/v_{\text{ter}}^2)$. Separating and using partial fractions,
$$\int\frac{dv}{1 - v^2/v_{\text{ter}}^2} = v_{\text{ter}}\operatorname{artanh}\frac{v}{v_{\text{ter}}} = gt,$$
so $v = v_{\text{ter}}\tanh(gt/v_{\text{ter}})$. The natural time scale is $v_{\text{ter}}/g$, the time the body would take to reach terminal speed if it fell freely all the way. For a skydiver that is about six seconds; the hyperbolic tangent reaches $0.76$ at one such time and $0.96$ at two.
Integrating the speed gives $y = (v_{\text{ter}}^2/g)\ln\cosh(gt/v_{\text{ter}})$. For large $t$, $\ln\cosh u \approx u - \ln 2$, so the body falls at terminal speed but trails a free-falling start by the fixed distance $(v_{\text{ter}}^2/g)\ln 2$.
Engineers write quadratic drag as $f = \tfrac{1}{2}C_d\rho Av^2$, where $\rho$ is the air density, $A$ the frontal area and $C_d$ a dimensionless drag coefficient measured in wind tunnels. A sphere has $C_d \approx 0.47$, which with $\rho = 1.29$ kg/m³ and $A = \pi D^2/4$ gives $c = 0.24\,D^2$, the $\gamma = 0.25$ of the last lesson.
A skydiver spread-eagle has $C_dA$ of about $0.4$ m² and head-down about $0.1$ m², so the terminal speed doubles, from near $55$ m/s to over $110$ m/s, when she tucks. A family sedan has $C_d$ around $0.3$, a pickup around $0.4$, and a road bicycle with rider about $0.9$ on an area of half a square meter. The coefficient hides the difficult fluid mechanics in one measured number, so the mechanics problem stays simple.
The drag is $-cv\,\vec{v}$, with magnitude $cv^2$ and direction opposite to $\vec{v}$. Its components are $-cv_x\sqrt{v_x^2 + v_y^2}$ and $-cv_y\sqrt{v_x^2 + v_y^2}$, so the horizontal deceleration depends on how fast the body is falling, and the other way round. Linear drag escaped this because $-b\vec{v}$ has components $-bv_x$ and $-bv_y$ with no cross terms.
The coupled equations are easy for a computer: step the velocity forward by $\Delta\vec{v} = (\vec{g} - (c/m)v\vec{v})\Delta t$ and the position by $\vec{v}\Delta t$, with a time step much shorter than $v_{\text{ter}}/g$. The results are striking. A batted baseball leaving at $45$ m/s and $35°$ would carry about $200$ m in a vacuum and carries about $120$ m in air, and it comes down more steeply than it went up.
Checking an answer. Early on, $\tanh u \approx u$ and $\ln\cosh u \approx u^2/2$, so the fall must start as $v = gt$ and $y = \tfrac{1}{2}gt^2$. Late, the speed must approach $v_{\text{ter}}$ from below. A terminal speed must scale as $\sqrt{m}$ and as $1/D$ for a fixed mass. Units: $c$ is kg/m, so $mg/c$ is m²/s² and its root is a speed.
For a sphere of density $\rho_s$, the mass is $\rho_s\pi D^3/6$ and $c = \gamma D^2$, so $v_{\text{ter}} = \sqrt{\rho_s\pi gD/6\gamma}$ grows as $\sqrt{D}$. A hailstone four times the diameter falls twice as fast. This is gentler than the $D^2$ of the linear regime, which is why raindrops of very different sizes fall at not very different speeds: a $1$ mm drop at about $4$ m/s, a $5$ mm drop at about $9$ m/s, where it flattens and breaks up.
Density enters the same way. A ping-pong ball and a golf ball of similar size differ in density by a factor of about fifteen, and their terminal speeds by about four: roughly $8$ m/s against $31$ m/s. That is why a ping-pong ball is hard to throw far however hard it is thrown, and why the lightest balls in any sport are the ones whose flight is dominated by the air.
When the ratio $1.6 \times 10^3\,Dv$ is near one, neither term can be dropped. The one-dimensional equation $m\dot{v} = mg - bv - cv^2$ is still separable, with terminal speed the positive root of $cv^2 + bv - mg = 0$, and its solution is a hyperbolic tangent shifted and rescaled. Small raindrops of about $0.1$ mm, and dust in a gust, live in this regime.
In practice engineers cover the whole range with a drag coefficient $C_d$ that depends on the Reynolds number, read from a measured curve: large at low Reynolds number, where it reproduces Stokes's law, flat near $0.47$ over a wide middle range, and dropping sharply near $\text{Re} = 3 \times 10^5$, the drag crisis, which is why a golf ball's dimples, which trigger it early, let the ball fly farther.
Skydivers at drop zones in Ohio exit at about $4000$ m. A spread-eagle jumper of $80$ kg reaches about $56$ m/s, near $125$ miles per hour, and gets to 90 percent of it in about eight seconds and $270$ m of fall. The next minute of free fall is at an almost steady speed, which is what makes formation skydiving possible: every jumper is at terminal speed and can adjust it by changing posture.
Opening the parachute raises $C_dA$ from about $0.4$ m² to about $25$ m², sixty times more, so the terminal speed falls by $\sqrt{60} \approx 8$, to about $7$ m/s, a landing like jumping from a wall two and a half meters high. The deceleration peaks at a few times $g$ as the canopy opens, set by how quickly it inflates, which is why modern parachutes use a slider that slows the opening to spread the braking over a second or two.
Hail forms in the updrafts of strong thunderstorms, and the Great Plains, from Texas to the Dakotas, see more of it than anywhere else in the country. A hailstone grows while the updraft holds it aloft, and falls when its terminal speed exceeds the updraft's speed. A $1$ cm stone needs only about $14$ m/s to fall; a $5$ cm stone, the size of a golf ball, falls at about $30$ m/s, so storms that make it have updrafts of that strength.
The kinetic energy at impact grows fast with size: mass grows as $D^3$ and speed squared as $D$, so energy grows as $D^4$. A golf-ball stone carries about $25$ J, enough to dent a car and crack a shingle, and a baseball-sized stone of $7$ cm carries four times that. The National Weather Service calls a storm severe when its hail reaches $2.5$ cm, and insurers in Colorado and Texas pay billions of dollars a year for roofs and cars damaged by it.
With linear drag, $v_{\text{ter}} = mg/b$ is proportional to the mass, and that rule is easy to carry over. With quadratic drag, $v_{\text{ter}} = \sqrt{mg/c}$: four times the mass only doubles the terminal speed. The same square root is why a heavy skydiver and a light one, side by side, fall at nearly the same speed, and why jumpers of different weights adjust their body position rather than their weight to fly together.
A related error is to separate the $x$ and $y$ equations of a thrown ball as if the drag were linear. The quadratic drag on the horizontal motion depends on the vertical speed, so the two cannot be solved one at a time.
A baseball has $m = 0.145$ kg and $D = 0.070$ m. Find its quadratic coefficient.
$c = \gamma D^2 = 0.25 \times 0.070^2 = 1.23 \times 10^{-3}\ \text{kg/m}$
$c = \gamma D^2$ for a sphere in air.
Balance drag against weight.
$cv_{\text{ter}}^2 = mg$
Zero net force at terminal speed.
Solve for the terminal speed.
$v_{\text{ter}} = \sqrt{\dfrac{0.145 \times 9.8}{1.23 \times 10^{-3}}} = \sqrt{1160}$
Substitute the mass and coefficient.
Evaluate the terminal speed.
$v_{\text{ter}} \approx 34\ \text{m/s}$
About $76$ miles per hour.
Compare with a fastball.
$v_{\text{pitch}} \approx 42\ \text{m/s} > v_{\text{ter}}$
At pitching speeds the drag exceeds the ball's weight, so drag is never negligible in baseball.
A rider and bicycle, $m = 80$ kg with $c = 0.20$ kg/m, stop pedaling at $10$ m/s on level road. Find the time constant.
$\tau = \dfrac{m}{cv_0} = \dfrac{80}{0.20 \times 10} = 40\ \text{s}$
Ignoring rolling resistance.
Find the speed after $40$ s.
$v = \dfrac{10}{1 + 40/40} = 5.0\ \text{m/s}$
One time constant halves the speed.
Find the distance covered by then.
$x = v_0\tau\ln 2 = 10 \times 40 \times 0.693 = 277\ \text{m}$
$x = v_0\tau\ln(1 + t/\tau)$.
Find the speed after $120$ s.
$v = \dfrac{10}{1 + 3} = 2.5\ \text{m/s}$
Three time constants.
Find the distance after $120$ s.
$x = 400\ln 4 = 555\ \text{m}$
Twice the distance at $40$ s: $\ln 4 = 2\ln 2$.
A skydiver of $80$ kg falls spread-eagle with $c = 0.25$ kg/m. Find her terminal speed.
$v_{\text{ter}} = \sqrt{\dfrac{80 \times 9.8}{0.25}} = \sqrt{3136} = 56\ \text{m/s}$
About $125$ miles per hour.
Find the natural time scale.
$\dfrac{v_{\text{ter}}}{g} = \dfrac{56}{9.8} = 5.71\ \text{s}$
The time to reach $v_{\text{ter}}$ in free fall.
Evaluate the argument at $t = 5.0$ s.
$\dfrac{gt}{v_{\text{ter}}} = \dfrac{9.8 \times 5.0}{56} = 0.875$
Dimensionless.
Find her speed at $5.0$ s.
$v = 56\tanh(0.875) = 56 \times 0.704 = 39.4\ \text{m/s}$
Free fall would give $49$ m/s.
Find how far she has fallen.
$y = \dfrac{56^2}{9.8}\ln\cosh(0.875) = 320 \times \ln 1.408 = 320 \times 0.342 = 109\ \text{m}$
Free fall would give $122$ m.
Check the early behavior.
$\tanh u \approx u \Rightarrow v \approx gt \text{ for } t \ll 5.71\ \text{s}$
At first she falls freely, as she must.
Write the terminal speed.
$v_{\text{ter}} = \sqrt{\dfrac{mg}{c}}$
Drag balances weight.
Substitute the values.
$v_{\text{ter}} = \sqrt{\dfrac{0.20 \times 10}{2.0 \times 10^{-3}}} = \sqrt{1000}$
SI units.
Evaluate the terminal speed.
A ball falling with quadratic drag has terminal speed $46$ m/s. A second ball of the same size and shape, so the same $c$, has four times the mass. What is its terminal speed?
Complete the worked solution: a cart of mass $720$ kg rolls on a level track at $16$ m/s, slowed only by quadratic drag with $c = 5$ kg/m. Find the time constant $\tau = m/cv_0$ in seconds, the speed after one time constant, and the speed after three time constants, in m/s.
Divide the mass by the product of the coefficient and the launch speed.
$\tau = \dfrac{m}{cv_0} =$ k
The time for the speed to halve.
Set the time equal to one time constant.
$v(\tau) = \dfrac{v_0}{1 + 1} =$ h
The denominator has doubled.
Set the time equal to three time constants.
$v(3\tau) = \dfrac{v_0}{1 + 3} =$ q
Quadratic drag slows the cart ever more gently as it slows.
Check the pattern of the speeds.
$\text{the reciprocal speed rises by equal steps in equal times}$
$1/v$ grows linearly, so the second halving takes twice as long as the first.
Match each quantity for motion with quadratic drag to its formula.
| $\sqrt{mg/c}$ | $v_0/(1 + t/\tau)$ | $v_{\text{ter}}\tanh(gt/v_{\text{ter}})$ | $\gamma D^2$ | |
|---|---|---|---|---|
| terminal speed | ||||
| horizontal coasting speed | ||||
| vertical speed from rest | ||||
| quadratic coefficient |
A skydiver falling with quadratic drag has terminal speed $20$ m/s. Fill in the terminal speed, in m/s, for four times the mass, for four times the drag coefficient, and for nine times the mass.
| $v_{\text{ter}}$ (m/s) | |
|---|---|
| four times the mass | |
| four times the coefficient | |
| nine times the mass |
A body launched horizontally at $7$ m/s feels only quadratic drag, with $\tau = m/cv_0 = 5$ s. Write its speed $v$ as a formula in the time $t$, in seconds.
Answer:
A hailstone is an ice sphere $1$ cm across with mass $0.4712$ g. Taking $\gamma = 0.25$ N s²/m⁴ and $g = 9.8$ m/s², what is its terminal speed, in m/s?
Answer: m/s for the hailstone
A skydiver of mass $70$ kg, with gear, jumps from a plane over Ohio and falls spread-eagle with $c = 0.25$ kg/m. Her terminal speed is $\sqrt{mg/c} = 52.38$ m/s. Taking $g = 9.8$ m/s², how many seconds after leaving the plane does she reach 90 percent of it?
Answer: s to 90 percent of terminal
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A body launched horizontally at $8$ m/s feels only quadratic drag, with $\tau = m/cv_0 = 3$ s. Write its speed $v$ as a formula in the time $t$, in seconds.
Answer:
You can solve motion with quadratic drag. Explain to someone why a skydiver twice as heavy does not fall twice as fast.
18. Your turn: a sphere with $c = 2.0 \times 10^{-3}$ kg/m has mass $0.20$ kg. Find its terminal speed with $g = 10$ m/s²., step 3
$v_{\text{ter}} \approx 31.6\ \text{m/s}$
The square root of a thousand.