# deriving a differential equation for projectile motion

We have a projectile, which we will consider spherical, of radius $$r$$ density $$\rho_m$$ and mass $$m$$, and we begin by firing it from a point $$(0,h_0)$$ at an initial velocity $$v_0$$ at an angle $$\theta_0$$. From this we can say that: $$x'(0)=v_0\cos(\theta_0),\,y'(0)=v_0\sin(\theta_0)\tag{1}$$ If we neglect all forces acting except those due to gravity then we can say: $$x''=0,\,y''=-g\tag{2.1}$$ Integrating to give: $$x'=C_1,\,y'=-gt+C_2\tag{2.2}$$ $$x=C_1t+C_3,\,y=-\frac{g}{2}t^2+C_2+C_4\tag{2.3}$$ Now by applying initial conditions we get: $$x=v_0\cos(\theta_0)t\tag{2.4.1}$$ $$y=-\frac g2t^2+v_0\sin(\theta_0)t+h_0\tag{2.4.2}$$ However, I would like to do the same thing as this but not neglecting air resistance (and if possible I would like to include the effect of spin). My thoughts to doing this would be: If $$F_x,F_y$$ denote force in the $$x$$ and $$y$$ directions, then we can say: $$\sum F_x=mx''\,\,,\,\,\sum F_y=my''\tag{3.1}$$ I have found that for an object moving through air, drag is equal to: $$F_{drag}=\frac 12CA\rho_{f}v^2\tag{3.2}$$ Since this is the total velocity of the object in question, we would have to say: $$v=\sqrt{(x')^2+(y')^2}\tag{3.3}$$ We now know the total drag force but obviously need the force in each direction. I believe the drag would always act in the opposite direction to motion. Because of this I would say: $$x''=-\frac{CA\rho_f}{2m}\left[(x')^2+(y')^2\right]\cos(\theta)\tag{3.4.1}$$ $$y''=-\frac{CA\rho_f}{2m}\left[(x')^2+(y')^2\right]\sin(\theta)-g\tag{3.4.2}$$ For some reason I can't visualise how I would get this $$\cos\theta$$ and $$\sin\theta$$ in terms of $$x$$ and $$y$$, maybe it would be clearer if I converted to vectors? I'm not sure how to continue from here to get a differential equation for $$x''$$ and $$y''$$. Thanks

EDIT:1

After getting help from users I end up with the differential equation: $$\frac{d^2}{dt^2}\begin{pmatrix}x\\y\end{pmatrix}=-\frac{CA\rho_f}{2m}\sqrt{(x')^2+(y')^2}\frac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix}-\begin{pmatrix}0\\g\end{pmatrix}$$

Notice that the drag force is always opposite to velocity. You can then write $$\cos\theta=\frac{v_x}v=\frac{x'}{\sqrt{x'^2+y'^2}}\\\sin\theta=\frac{v_y}v=\frac{y'}{\sqrt{x'^2+y'^2}}$$

• Exactly what I was looking for. Thank you! – Henry Lee May 27 at 18:33