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I'm trying to find the equation for a ball thrown from the ground with an initial velocity. Are these differential equations correct? I solved these and set the integrating constant to $v_0cos(\theta)$ for $v_x$ and $v_0sin(\theta)$ for $v_y$ and integrated again to get the function of position. Is that the correct approach?

$\dot{v_x}=\frac{k}{2m}v_x^2$ and $\dot{v_y}=\frac{k}{2m}v_y^2-g$ where $k:= C_d\rho A$

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I assume that you are taking the drag force as having magnitude $k|{\bf v}|^2$? In that case the you need to resolve the components of the drag vector which points backwards along ${\bf v}$, and get $$ m\dot v_x= -kv_x\sqrt{v_x^2+v_y^2}, $$ $$ m\dot v_y= -kv_y\sqrt{v_x^2+v_y^2}-g. $$ These are essentially impossible to solve analytically in general, but for a near horizontal trajectory (a rifle bullet say) we can use the Siacci approximation, in which we ignore the effect of $v_y$ under the square root.

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