For reference, I'm using this equation for drag

$$d = c \times {\rho \times v^2 \over 2} \times a$$

And this one for finding terminal velocity

$$v = \sqrt{2 \times m \times g \over \rho \times a \times c}$$

For the purposes of this question, you can assume $c$, $\rho$, $a$, and $m$ are all $1$ and the object we are dealing with is a sphere. These two equations then simplify to

$$d = {v^2 \over 2}$$ $$v = \sqrt{2 \times g}$$

Now imagine $g=50$. The terminal velocity of the object in this environment would be $v = \sqrt{2 \times 50} = 10$. This object then has a velocity vector of $(0, -10)$. It's experiencing a downward force of $50$ units due to gravity, and drag is pushing it upwards by $d = {10^2\over 2} = 50$ units, perfectly cancelling out gravity, thus it is at its terminal velocity.

Now imagine a sideways force being applied. Its velocity vector now becomes $(4, -10)$. Calculating for drag, we can see the drag force has a magnitude of ${\sqrt{4^2 + (-10)^2}^2 \over 2} = 58$ units.

The unit vector in the direction of the velocity is given by

$$\theta = \text{atan2}(-10, 4)$$ $$U = (\cos \theta, \sin \theta)$$

And since drag is applied in the opposite direction of motion, we know the drag vector is $D = d \times -U$. And after actually crunching the numbers for the Y-component of $D$, we see that the object is experiencing an upwards force of $53.85...$ units, which is more than the downward force of gravity.

So, my question is, what's going on here? Does the sideways motion actually truly affect the vertical motion? Am I missing something in the calculations? Intuition would tell me that no matter how fast an object is moving sideways (ignoring things like lift, we're dealing with a sphere) its vertical motion will be unaffected. Is that intuition wrong?

Granted, my calculations are fundamentally 2D, but imagining similar situations in 3D gives the same result. The drag vector still has a stronger upwards component, which still raises the same questions.


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