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Due to missing a lecture (for health reasons), I'm reading my lecturer's notes on the resisted motion of a particle where air resistance is proportional to the square of the particle's speed.

Here is what I gathered and understood well:

If a particle is fired vertically upwards, where the resistance is $\space mkv^2 \space$, in which $\space k \space$ is a constant and $\space v \space$ is the speed. We can express the equation of motion as:

$m \ddot{y} = -mk{v^2} -mg$

Hence:

$\ddot{y} = -k{v^2} -g$

Now here is where I get confused. The notes then proceed to say:

Had the particle been falling under gravity, the RHS would be:

$mg-mkv^2$

Taking $\space y \space$ as the distance fallen. The acceleration would be zero when:

$g=kv^2$

OR

$v=\sqrt{g \over k}$

This is the limiting speed usually written as $\space c \space$. Therefore $\space c=\sqrt{g \over k} \space$. Thus telling us that:

$k={g\over c^2}$

We can use this in the current model like so:

$\ddot y = -g - {gv^2 \over c^2} = -g\big( {c^2 + v^2\over c^2} \big)$

But why are we using it in the model?

Is that not assuming the particle is falling under gravity, when it's not? It's stated that the particle has been fired UPWARDS.

It's like we changed the rules to fit our situation. Is there something I'm not understanding?

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Max Echendu
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1 Answers1

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That speed is just the "limiting speed", which everyone I've ever met refers to instead as "terminal velocity" but sure. All that ended up happening is that your equation was re-scaled to account for terminal velocity, but everything else still works out (try it!). It puts your equation into a form where something that is universally true for all such masses holds (namely, the terminal velocity), and I suspect it will let you work with this terminal velocity to do other things later.

Michael Stachowsky
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