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Given the initial velocity $v_0$ and angle $\theta$ of a projectile on the ground, using Newton's second law and the acceleration due to gravity $\mathbf g=\left\langle0,-g\right\rangle$, I was able to derive its position vector function:

$$ \mathbf F=m\mathbf a=m\mathbf g\implies\mathbf r(t)=\left(v_0t\cos\theta,-\frac g2t^2+v_0t\sin\theta\right). $$

I now want to introduce drag into this function. From my differential equations book, I was able to deduce that

$$ m\mathbf a+c\mathbf v=m\mathbf g, $$

Where $c$ is some scalar. Is this correct? If so, how can I go about solving for $\mathbf r$? Integrating factors make no sense to me in this context.

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  • $\begingroup$ @Bye_World, that's pretty cryptic. As $v$ has time derivatives of the position functions $x(t)$ and $y(t)$, it might be better if you could be more explicit. $\endgroup$
    – Simon S
    Nov 27, 2014 at 2:48
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    $\begingroup$ Josué, you can find a discussion here: farside.ph.utexas.edu/teaching/336k/Newtonhtml/node29.html $\endgroup$
    – Simon S
    Nov 27, 2014 at 4:05
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    $\begingroup$ I'd recommend to use normal parenthesis, ( ), instead of \angles when working with vectors, since they're almost universally used for inner products. $\endgroup$
    – hjhjhj57
    Nov 27, 2014 at 7:31
  • $\begingroup$ An issue with generic drag equations: forces like drag are non-conservative and act only in opposition to the direction of motion. Thus, writing an equation like you do is only useful when you know the direction of motion is not changing (e.g., analyzing a falling object). Something like $-k |\mathbf{v}|$ in the equation may be closer to reality. $\endgroup$
    – Ron Gordon
    Nov 27, 2014 at 8:03

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I think (I can't actually remember at the moment) that the dragging factor is certainly proportional to the speed. So, the last equation you wrote yields two ordinary differential equations: \begin{align*} &\frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = 0\\ &\frac{d^2}{dt^2}r_y(t) + \frac{c}{m}\frac{d}{dt}r_y(t) + g = 0 \end{align*} To solve the first one, just observe that: $$ \frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = \frac{d}{dt}\left(\frac{d}{dt}r_x(t) + \frac{c}{m}r_x(t) \right)=0, $$ which means that $\frac{d}{dt}r_x(t) + \frac{c}{m}r_x(t)$ is constant in time. Do you know how to solve this?

To solve the second equation, observe that it's the same as the first one but with the constant term added. Remember that the general solution of an inhomogeneous ODE, is the sum of the general solution and one particular solution (for example, $r_y(t) =-\frac{gm}{c}t$ is a particular solution).

Let me know if something isn't clear.

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  • $\begingroup$ I see. So I have to treat the components individually. $\endgroup$
    – wjmolina
    Nov 27, 2014 at 18:03
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    $\begingroup$ of course, physically each component is independent. $\endgroup$
    – hjhjhj57
    Nov 27, 2014 at 18:14

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