A skydiver's velocity after jumping from an initial height can be modeled by the equation

$$ m \frac{dv}{dt}=mg-k v^2. $$

I am trying to solve this analytically given the initial condition $v(0)=0$, but don't know how. It's not doable by integrating factor or Bernoulli's substitution, nor is it exact. While it is separable, the standard trig sub $v=\sqrt{mg/k} \sec(\theta)$ doesn't make sense for the initial condition. If someone could give me the initial substitution, I'd be really grateful.


  • $\begingroup$ I think you are misusing the identity $1 + \tan^{2}(x) = \sec^{2}(x)$. $\endgroup$ – user1337 Mar 20 at 0:47
  • $\begingroup$ You're right, I think I had been working too long $\endgroup$ – ChrisWong Mar 21 at 1:07

To start with, I would recommend to separate the variables:

\begin{align*} m\frac{\mathrm{d}v}{\mathrm{d}t} = mg - kv^{2} \Longleftrightarrow \int\frac{\mathrm{d}v}{mg - kv^{2}} = \int\frac{\mathrm{d}t}{m} \end{align*} Then we can make the substitution $\displaystyle v = \sin(\theta)\sqrt{\frac{mg}{k}}$. Consequently, one has that \begin{align*} \int\frac{\mathrm{d}v}{mg-kv^{2}} & = \sqrt{\frac{1}{mgk}}\int\frac{\cos(\theta)}{1-\sin^{2}(\theta)}\mathrm{d}\theta\\\\ & = \sqrt{\frac{1}{mgk}}\int\frac{\mathrm{d}\theta}{\cos(\theta)}\\\\ & = \sqrt{\frac{1}{mgk}}\ln|\sec(\theta) + \tan(\theta)| + C \end{align*}

If we denote $(mg/k)^{1/2}$ by $a$, we have that \begin{align*} \sec(\theta) + \tan(\theta) & = \frac{1+\sin(\theta)}{\cos(\theta)} = \frac{1+a^{-1}v}{\sqrt{1-a^{-2}v^{2}}} = \frac{a + v}{\sqrt{a^{2} - v^{2}}} = \sqrt{\frac{a+v}{a-v}} \Longrightarrow\\\\ \int\frac{\mathrm{d}v}{mg - kv^{2}} & = \frac{1}{2}\sqrt{\frac{1}{mgk}}\ln\left|\frac{a+v}{a-v}\right| = \frac{t}{m} + C \end{align*} Since $v(0) = 0$, we conclude that $C = 0$, and we are done.

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