# Air drag on a projectile [closed]

A projectile moving through two spatial dimensions has a resolution of two velocities, $$\vec i(V_{0}\cos\theta)+\vec j(V_0\sin\theta -gt)$$ each of which should have their own air drag(?) But also orthogonal to each other, just as the velocoties are. Is it correct to sum the two air perpendicular air drags as: $$\sum F_{drag}= \vec i(bV_0\cos\theta)+\vec j(bV_0\sin\theta-gtb)$$

where $$b$$ is supposed to represent the product of the drag coefficient, fluid density, etc. (using the Stokes model of proportionality)

seems like sensible reasoning (even if I do say so myself hehe) but i still unsure. googling could not reveal :( so I beseech thee.

• I think he wants to know if you can apply the drag formula separetely to the vertical and horizontal velocity components, and add the results, of if you need to apply it to the resultant speed first, and only then separate the components. – ksousa May 31 at 21:42
• hello, my Q was about the correctness of the posted reasoning (about the air drags in the $\vec i$ and $\vec j$ directions being perpendicula)r and then if it is correct to write their sum as $$F_{drag}=\vec i(bV_0\cos\theta)+\vec j(bV_0\sin\theta-gtb)$$ – dia bu May 31 at 21:43

Your suggestion is correct in someways but incorrect in others. A better representation of the drag force would be

$$\vec{F}_{drag} = \alpha (V_x(t)\space\vec{i}\space +\space V_y(t)\space\vec{j})$$ Where $$V_{x,y}$$ are simply the time dependent velocities of the projectile. You will have to use this to set up differential equations in both the $$x$$ and $$y$$ directions in this case, and solve them subject to suitable boundary conditions (i.e. $$V_x(0)=V_0)$$ or something to that effect.

One of the issues with your formulation is that you have included the change in velocity due to gravity in your y-velocity, which makes it seems as though you believe your formula works at all times $$t$$. However, you have to remember that the drag force will alter the velocities as a function of time, so this doesn't work. Explicitly, what this means is that your initial two dimensional velocity decomposition is invalid.

It's also worth noting that the drag force is often proportional to velocity squared, instead of velocity, so maybe you should look at whether or not this is a more appropriate model for your system. In this case the formula above is the same, but you just change it to squared velocities, and the constant $$\alpha$$ will be different.

• thank this was helpful, now i see that my post's velocity resoultion was incorrect because it fails to take into account the drag force(which is the whole point LOL)e SO . with the goal of an autonomous ODE to solve for a $V(t)$ function in mind me think i should forget about writing just the drag force and write an equation for all of the forces on the body relying on constants of integration to account for things like initial velocity, so $$F=\vec i(\alpha V_x)+\vec j(mg+\alpha V_y)$$ , then i can divide by mass to form a first order, linear autonomous differential equation?or is wrong? – dia bu May 31 at 23:27
• Yeah, but you'll have two ODEs, one in $x$ and one in $y$. Also be careful with the signs, it shouldn't effect the results as long as you get the co-ordinates orientated in a suitable manner, but still worth being conscientious of. – MC2k Jun 1 at 9:00

I recommend treating components as a way to calculate, not a way to think. The linear drag force is $$-b\vec v$$: proportional to, and opposite to, the velocity. If you expand that into components, you get what you wrote, except that you’re missing a negative sign. Thinking that there are “two drag forces” is a confusing approach.

• I think this is somewhat close minded. In fact, in the absence of significant velocity/position dependent effects, I think one of the most powerful tactics in classical mechanics is transferring frames so that you can reduce the number of relevant dimensions. Indeed in Electromagnetism, with the Lorentz force, it is almost necessary to think in separate components, at least when initialising the problem. – MC2k May 31 at 22:07
• @MC2k I’m not sure what you’re talking about. One can, and in my opinion should, visualize $q(\vec E +\vec v\times\vec B)$ in a coordinate-free way. The whole point of vector notation is to think in terms of vectors, not their components. To me it is clear that thinking in components has only confused the OP. The key concept is that drag force is opposite to the velocity. – G. Smith May 31 at 22:19

The drag force $$\vec{F}_d$$ components are:

$$\vec{F}_d=-f(v)\,\vec{e}_v=-f(v)\,\frac{\vec{v}}{v}$$

where $$f(v)$$ is the drag law and v is the magnitude of the velocity vector $$\vec v$$

I) linear drag law : $$f(v)=b\,v$$

$$\vec{F}_d=-f(v)\frac{\vec{v}}{v}=-b\,\vec{v}=-b\,\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}$$

II) quadratic drag law : $$f(v)=c\,v^2$$

$$\vec{F}_d=-f(v)\frac{\vec{v}}{v}=-c\,v\,\vec{v}=-c\,\sqrt{v_x^2+v_y^2}\,\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}$$