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Suppose I have a ball moving through the air parallel to the ground, and diagonally in some 2D $x$ , $y$ coordinates (along the line $y=x$). Let's say its velocity is 1 units, which means on the $x$ axis and on the $y$ axis separately it's $\frac{\sqrt{2}}{2}$.

Modeling air resistance as approximately $F=Kv^{2}$, say with $K=1$, I could say that along the $x$ axis, the resistance $F_{x}$ is $K\frac{\sqrt{2}}{2}^{2}$ = 1, and similarly $F_y=1$, so the total air resistance, acting diagonally, is $F=\sqrt{2}$.

On the other hand, if I redraw coordinate axes so that one of them coincides with the velocity vector, along that axis $v=1$ , $F=Kv^{2}=1$.

Where did I go wrong? Is it impermissible to consider air resistance acting separately in each dimension, and why?

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    $\begingroup$ only linear air resistance behaves in the way you're proposing. $\endgroup$
    – AXensen
    Mar 29, 2023 at 14:34
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    $\begingroup$ how did you get $K(\frac{\sqrt{2}}{2})^2 = 1$ ? shouldnt it be $\frac{1}{2}$ $\endgroup$
    – Naveen V
    Mar 29, 2023 at 15:29

3 Answers 3

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As a vector, if you want the air resistance force to (a) have a magnitude proportional to the square of the speed and (b) directed opposite the velocity vector, then the correct formula would be $$ \vec{F} = - K |\vec{v}| \vec{v}. $$ It is not too hard to see that this satisfies the criteria above. But note that in terms of components, this means that \begin{align*} F_x &= - K ( v_x^2 + v_y^2)^{1/2} v_x \neq - K v_x^2, \text{ and} \\ F_y &= - K ( v_x^2 + v_y^2)^{1/2} v_y \neq - K v_y^2. \end{align*}

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  • $\begingroup$ Thank you. I think I understand now. My confusion originally stemmed from a fascinating problem where air resistance acts in two dimensions (the problem: A football at distance L from a gate is kicked straight at it with speed v>0, and due to a sidewise perpendicular wind of speed u>0 arrives at the gateline at distance h from the gate; find the time of travel to the gateline). It's natural to consider velocities and resistance forces separately in x and y, but beyond being difficult to solve with quadratic force, I now realize it's incorrect b/c the force acts along the total velocity vector. $\endgroup$ Apr 2, 2023 at 7:54
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You are applying your linear thinking to a nonlinear situation.

If a number is split into pieces, you can add up the pieces to recover the original number. If the number is raised to a power other than 1.0, that is no longer possible.

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Your math is incorrect.

You say for $K=1$

$$F_x= K \left(\frac{\sqrt{2}}{2}\right)^2 = 1$$

The correct expression is:

$$F_x= K \left(\frac{\sqrt{2}}{2}\right)^2 = (1)*\frac{1}{2} = \frac{1}{2}$$

Same for $F_y$. So the resultant drag force is:

$$\sqrt{F_x^2+F_y^2}=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} = \sqrt {\frac 1 2} $$

As a tip, $\frac{\sqrt{2}}{2} = \sqrt{\frac1 2 } \; \; $ and sometimes it is more instructive to write it that way.

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  • $\begingroup$ My math is wrong (thanks), but so is yours, I think. F_x=F_y=1/2, as you say, but then the resultant force is $\sqrt(F_x^2+F_y^2) = \sqrt(1/4 + 1/4) = \sqrt(2)/2.$ $\endgroup$ Apr 1, 2023 at 13:16

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