Motion of a falling object with air resistance

Question

I was wondering how you would model the velocity of a falling object, taking into account air resistance. Bear in mind I have only studied basic calculus, and have no experience with differential equations.

Considering a general solution for mass $m$ and drag $kv^2$, where $k=\frac{1}{2} \rho AC_D$. Let's say that when $t=0$, $v=0$.

Using Newton's Second Law: $mg - kv^2 = ma$, hence $a = g - \frac{k}{m} v^2$.

From here, I tried to integrate with respect to $t$, which gave $v = gt-\frac{kt}{m}v^2$. This, when solved for v, seems to model the velocity correctly, but I have been led to believe that solutions to differential equations of this kind will involve $e^x$ in some way. Is my solution correct, or have I integrated incorrectly? How do we go from here to a function for v in terms of t?

Answer 1

Remembering that $a=\dfrac{dv}{dt}$, $$m\frac{dv}{dt}=mg-kv^2$$ $$\frac{dv}{mg-kv^2}=\frac{dt}{m}$$ Integrating both sides, $$\int\frac{dv}{mg-kv^2}=\int\frac{dt}{m}$$ You can find an analytical solution from here. Here is the full solution on my blog if you are interested:https://philosophicalmath.wordpress.com/2017/10/21/terminal-velocity-derivation/

Answer 2

You should account for the fact that acceleration and velocity (as much as position) might be —and in fact are— varying with time.

So the equation you got from Newton's 2nd law means $$a(t)=g-\tfrac k m [v(t)]^2.$$

Then, integration in $t$ would imply

$$v(t)-v(t_0)=\int_{t_0}^t a(\tau) d\tau=\int_{t_0}^t g-\tfrac k m [v(\tau)]^2 d\tau.$$

Here you can see your mistake: you omitted the fact that $v$ is not constant in $t$.

Here is where differential equations come into play. Avoiding technicalities, we can see that your equation can be rewritten as $$\big(g-\tfrac k m [v(t)]^2\big)^{-1}v'(t)=1,$$ that is $$\frac{v'(t)}{\big(\sqrt{g}-\sqrt{\tfrac k m} v(t)\big)\big(\sqrt{g}+\sqrt{\tfrac k m} v(t)\big)}=1.$$ Then, if we use indefinite integration in $t$ $$\int\frac{v'(t)}{\big(\sqrt{g}-\sqrt{\tfrac k m} v(t)\big)\big(\sqrt{g}+\sqrt{\tfrac k m} v(t)\big)}dt=\int dt,$$ which gives an equation of the form $$A \ln\left(\sqrt{g}-\sqrt{\tfrac k m} v(t)\right)+B\ln\left(\sqrt{g}-\sqrt{\tfrac k m} v(t)\right)=t+C.$$

Constants $A$ and $B$ are the only numbers such that $$\frac{A}{\sqrt{g}-\sqrt{\tfrac k m} v(t)}+\frac{B}{\sqrt{g}+\sqrt{\tfrac k m} v(t)}=\frac{1}{\big(\sqrt{g}-\sqrt{\tfrac k m} v(t)\big)\big(\sqrt{g}+\sqrt{\tfrac k m} v(t)\big)},$$

while constant $C$ can be any real number in a general setting, but has a unique value (in terms of $v_0$) if we impose the condition $v(t_0)=v_0$. Finding all these constants and solving for $v(t)$ gets you the classic formula for this kind of trajectory, which will indeed include exponential functions.