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.

  • $\begingroup$ 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

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


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