1
$\begingroup$

The velocity of a falling object (or the derivative with respect to time of the distance traveled by the falling object) is given by

$$ \dot{s} = \frac{rm}{k}\left(\frac{e^{rt}-e^{-rt}}{e^{rt}+e^{-rt}}\right), $$

where $r = \sqrt{gk/m}$.

Here, $g$ is acceleration due to gravity, $m$ is the mass of the object, and $k$ is a constant that captures air resistance.

I need to integrate this to obtain an exact formula for $s$. Can anyone give me some guidance for how to do this? I'm not sure what strategy is appropriate.

The answer should come out to be

$$ s(t) = \frac{V^2}{g}\ln\left(\cosh\frac{gt}{V}\right), $$

where $V=\sqrt{\frac{mg}{k}}$ is the terminal velocity.

|cite|improve this question
$\endgroup$
  • $\begingroup$ You would obtain $s$ by integration, not differentiation. $\endgroup$ – Math1000 Mar 3 '17 at 22:36
  • 1
    $\begingroup$ Consider $$\tanh(rt)=\frac{e^{rt}-e^{-rt}}{e^{rt}+e^{-rt}}$$ $\endgroup$ – Mengchun Zhang Mar 3 '17 at 22:37
  • $\begingroup$ @dbliss Bingo :) you got exactly the right answer $\endgroup$ – Mengchun Zhang Mar 3 '17 at 23:18
1
$\begingroup$

The limiting velocity is the value of $\dot{s}$ as $t\rightarrow\infty$, so $V^2=\frac{rm}{k}$

Integrating the expression you have gives (assuming $v=0$ when $t=0$) $$s=\frac Vr\ln\cosh rt$$

But $\frac{V^2}{g}=\frac mk=\frac Vr$

Hence $$s=\frac{V^2}{g}\ln\cosh\frac{gt}{V}$$

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.