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If I search for air resistance on a falling object, I find texts stating that it is proportional to the velocity.

$$F=mg-kv$$

Is this accurate or approximate?

Some texts say that when the speed is high, the air resistance is proportional to the square of the speed.

I would like to know the exact differential equation for a thrown sphere.

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The equation for air resistance that you mentioned is the stokes law. $$F_{drag}=6\pi \eta r v \propto v$$ This equation is only valid when velocity is extremely low. Why is that so? Because, while deriving the stokes law we assume it smoothly moves around the ball and for that the velocity must be low for such "laminar" flow.

Now, say the velocity is high, the ball is moving too fast for air to move around it. Then it encounters $\rho v A$ mass of air per unit time. That is,

$$\frac{dm}{dt}=\rho v A$$

Let us reasonably assume that the ball gives the air a velocity $v'$ which intuitively from collisions should be proportional to v.

So the rate of momentum that air gains would be: $$F_{drag}=\frac{1}{2} C \rho A v^2$$

Where C is the drag coefficient.

From newtons third law, ball experiences an opposite force: $$\vec{F}_{drag}=-\frac{1}{2} C \rho A v^2 \hat{v}$$ Where $\hat{v}$ is the unit vector in the direction of the velocity.

Now how much is "too much" velocity, where is this transition point?

This is decided by the reynolds number: $${\displaystyle \mathrm {Re} ={\frac {\rho vL}{\mu }}}$$

For a sufficiently small Reynolds number (below a critical value usually around $2\cdot 10^5$), the flow is not turbulent and you can take it to be proportional to $v$ and if it is significantly higher than the critical value it would be proportional to $v^2$.

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