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The differential equation $ma=mg-kv^2$ is to describe the motion of a particle that is falling from a tall building, and the air resistance is proportional to the square of the velocity of the particle.

I solve the DE by integration using the initial condition $x=0$ when $t=0$ and the assumption that $g \ge kv^2$, and obtained $v=\sqrt{\dfrac{g}{k}(1-e^{-2kx})}$, where $x$ is the distance the particle travels downwards in metres.

Did I obtain the correct particular solution?

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  • $\begingroup$ Why not check by plugging it back in? $\endgroup$ – The Count Sep 13 '18 at 13:51
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    $\begingroup$ You know what,I could have done that! Thank you. $\endgroup$ – Will Kim Sep 13 '18 at 13:52
  • $\begingroup$ You can post an answer to your own question, perhaps! $\endgroup$ – The Count Sep 13 '18 at 13:54
  • $\begingroup$ At first the answer given below seems odd as it is not immediately apparent what happens when $k=0$ reverting to the case where there is no air resistance, with the velocity at a given $x$ then independent of $m$. However in the limit $k \to 0$, $e^{-\frac{2k x}{m}} \approx 1- \frac{2kx}{m}$, with $v$ being as expected $v=\sqrt{2gx}$ $\endgroup$ – James Arathoon Sep 13 '18 at 18:03
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$$ma=mg-kv^2$$ $$mv\frac{dv}{dx}=mg-kv^2$$ $$\int mv\frac{dv}{mg-kv^2}=\int dx$$ $$-\dfrac{m\ln\left(\left|kv^2-gm\right|\right)}{2k}+C=x$$

at $x=0$ $v=0$ $$C=\dfrac{m\ln\left(\left|gm\right|\right)}{2k}$$

$$-\dfrac{m\ln\left(\left|kv^2-gm\right|\right)}{2k}+\dfrac{m\ln\left(\left|gm\right|\right)}{2k}=x$$

We have $mg\ge kv^2$

thus $$v^2=\frac{mg}{k}(1-e^{-2kx/m})$$

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  • $\begingroup$ Dont we always have $g \ge kv^2$? I simplified the solution further based on this assumption. $\endgroup$ – Will Kim Sep 13 '18 at 14:19
  • $\begingroup$ we always have $$mg\ge kv^2$$ $\endgroup$ – Deepesh Meena Sep 13 '18 at 14:19
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I solved the differential equation and came up with $$v=\sqrt{\dfrac{mg}{k}(1-e^{-2kx/m})}$$ Which is slightly different from your answer.

Are you assuming $m=1$?

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  • $\begingroup$ At the start, I divided the DE by m and let k be k/m. So in my answer I should really have k’ where k’=k/m. $\endgroup$ – Will Kim Sep 13 '18 at 20:58

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