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I have the following equations

$$\ x′′(t)=−\frac km x′(t)$$

$$\ y′′(t)=−\frac km y′(t) - g$$

where $k$ is the drag, $m$ is the mass of the object and $t$ is the time. $g$ is the gravity constant.

After integrating the functions twice, I end up with the following equations

$$\ x(t) = C_1 e^{\frac {-kt} {2}} + C_2$$ $$\ y(t) = C_3 e^{\frac {-kt} {2}} + C_4 - \frac {gmt}{k}$$

How should I determine the constants $C_1$, $C_2$ etc? I want to be able to set the angle $\alpha$ and velocity $v_0$.

For example $y'(0) = 30 \sin(\alpha)$

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closed as off-topic by Brandon Enright, Kyle Kanos, jinawee, Qmechanic May 12 '14 at 23:45

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  • $\begingroup$ First check your integrations, you have mistakes on that. And just some pedantic observation: you should have more integration constants, but it looks like that you already applied some not mentioned initial conditions. $\endgroup$ – Ignacio Vergara Kausel May 12 '14 at 15:53
  • $\begingroup$ @Ignacio Sorry forgot two signs. More than four integration constants in total? $\endgroup$ – mouduor May 12 '14 at 15:59
  • $\begingroup$ Should the first term in your second differential equation depend on $y'(t)$ rather than $x'(t)$? $\endgroup$ – BMS May 12 '14 at 16:05
  • $\begingroup$ @BMS Yes, now I hope everything is in order. :) $\endgroup$ – mouduor May 12 '14 at 16:10
  • $\begingroup$ There are 4 unknowns $C_1,C_2,C_3,C_4$, and there are 4 initial conditions $x(0)=A,y(0)=B,x'(0)=C,y'(0)=D$. So you have all the info you need to determine the $C_k$. $\endgroup$ – DumpsterDoofus May 12 '14 at 16:17
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I think this problem is sufficiently general to merit writing an explicit answer. My preference for formalizing the initial conditions are:

$$ x(0) = 0, y(0) = 0 \\ x'(0) = v_x, y'(0) = v_y \\ $$

I believe the solutions are:

$$ x(t) = -\frac{v_x m}{k} \left( {{\rm e}^{-{\frac {kt}{m}}}}-1 \right) \\ y(t) = - \frac{m}{k^2} \left( \left( gm+{\it v_y}\,k \right) \left( {{\rm e}^{-{\frac {kt}{m}}}}-1 \right) +gtk \right) $$

You can put things in terms of the initial velocity and launch angle if you desire. That should not be difficult.

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