# Free falling object with air resistance, what's the velocity as a function of time?

So, I'm looking into how velocity can be represented as a function of time when the air resistance is also function of velocity. The force of drag is assumed to be quadratic

$$\vec{F}_d = -C||\vec{v}||\vec{v}$$

where $$C$$ is an arbitrary constant depeding on the object's shape etc.

Obviously, we are assuming a gravitational force of $$-mg$$ near the surface of the Earth.

What confuses me is that this force itself is not constant to begin with. I have a sense that I can integrate my way out of this, so I'd really appreciate who could tell me where to start.

• I don't think you can integrate your way out of this one. Set it up as a numeric simulation with small time intervals. Plot the data, and obtain a best fit with a 4th or 5th order polynomial. Feb 11, 2016 at 3:36
• But because this is homework, it has to be accomplished algebraically. Feb 11, 2016 at 4:03
• What is the problem with using $m\dfrac{dv}{dt} = mg - Cv^2$?? You will get velocity as a function of time by this... Feb 11, 2016 at 4:41
• Terminology: If there is air resistance an object is by definition not free falling. Aug 30 at 8:11

You have a net force $$F = F_G - F_D$$ $$ma = -mg + C_0 v^2$$

Dividing both sides by m, letting $C_1 = C_0 / m$, $$a = -g + C_1 v^2$$ $$\frac{dv}{dt} = -g + C_1 v^2$$ $$\frac{dv}{-g+C_1 v^2} = dt$$

Now to integrate both sides, we need to use the limits $[0,t]$ for $dt$, $[v_0,v]$ for $dv$ (where $v_0$ is our initial velocity. $$\int_{v_0}^{v} \frac{dv}{-g+C_1 v^2} = \int_{0}^{t} dt$$

Plugging through Wolfram Alpha I get $$t = \frac{1}{\sqrt{g C_1}}\cdot[tanh^{-1}(\sqrt{\frac{g}{C_1}}v)-tanh^{-1}(\sqrt{\frac{g}{C_1}}v_0)]$$ which rearranges to $$v = \sqrt{\frac{C_1}{g}}\cdot tanh[\sqrt{g C_1}t + tanh^{-1}(\sqrt{\frac{g}{C_1}}v_0)]$$

And hence we have velocity as a function of time.

• Nice! Wolfram Alpha really came through on this one. And as Bill N pointed out, the v integral is DEFINITELY not trivial! Feb 11, 2016 at 12:10
• You really should have let diddlydoo actually do the work. He didn't ask for the answer to his assignment, he asked " I'd really appreciate who could tell me where to start." Feb 11, 2016 at 18:19

If it's falling only then you have $F_d=+Cv^2$, where up is the positive direction. You said there is a gravitational force $F_g=-mg$. Write a Newton's 2nd Law equation, set $a=\frac{dv}{dt}$, rearrange, to get dv/g(v) = dt, (I'll let you find $g(v)$) and integrate away. The $v$ integral is not trivial. Look it up in an integral table, if your teacher will let you do that.

Then you will have to invert the result to solve for the velocity function. That will take some algebraic work.

• The integral to find $v$ can be done using trigonometric substitution. To find $x$ you can use the formula for integral of hyperbolic functions. Feb 11, 2016 at 4:50
• Yo here we have Bill Nye the Science Guy! Feb 11, 2016 at 7:11

Thank you. I found this very useful indeed. However, I believe that there is an error in the formula for t - in particular in the arguments to the two atanh terms. The first atanh term should read atanh(sqrt(C1/g)*v). Similarly, the second should read atanh(sqrt(C1/g)*v0). Sorry, but I don't know how to do the fancy formatting. A simple units (SI) or dimensional analysis shows that there's something fishy about the given formula. sqrt(g/C1) has units of m/s and so represents a characteristic velocity, which v and v0 must be divided by - not multiplied by. The formula for V needs to be corrected commensurately. The best thing for me about this exercise is that I had to revisit my physics and maths of the 1960's - way too much IT in the interim!

By the way, it's worth noting that Integral(dx/(1-x^2)) = atanh(x).

From this it follows that Integral(dv/(g-C1.v^2)) = (1/sqrt(g*C1)) * atanh(sqrt(C1/g)*v)), which leads to the correct formula for t, etc.

• This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review Aug 30 at 8:20
• It corrects a wrong answer. Isn't that useful? That's all I'm interested in doing. Bye. Aug 30 at 10:11
• Then you should post it as comment under the answer containing the wrong elements. If you really want to make a post in the answer section, then your answer should be self-contained (and not depend on another answer, which may be deleted in the future). Aug 30 at 14:02