# Vertical Projectile with linear air drag

If a spherical ball of mass $$m$$ is thrown upwards with an initial velocity $$v_0$$ and it experiences linear drag, how do we come up with expressions for $$v(t)$$ using Newton's second law (taking up as the positive direction)?

During the ball's upward motion, the equation of motion can be given by:

$$m\dot{\vec{v_y}}=m\vec{g}+\vec{f}$$

Where $$\vec{f}$$ is the force due to air resistance.

$$\implies m\dot{v_y}\hat{y}=mg\hat{g}+f(v)\hat{f}$$

Now, here, we know that $$\hat{g}=-\hat{y}$$ and $$\hat{f}=-\hat{v}=-\hat{y}$$

$$\implies m\dot{v_y}\hat{y}=-mg\hat{y}-f(v)\hat{y}$$

Now, since we are dealing with linear drag, we have $$f(v)=bv$$:

$$\implies m\dot{v_y}=-mg-bv$$

During the ball's downward motion, the equation of motion can be given by:

$$m\dot{v_y}\hat{y}=mg\hat{g}+f(v)\hat{f}$$

Now, here, we know that $$\hat{g}=-\hat{y}$$ and $$\hat{f}=-\hat{v}=\hat{y}$$

$$\implies m\dot{v_y}\hat{y}=-mg\hat{y}+f(v)\hat{y}$$

$$\implies m\dot{v_y}=-mg+bv$$

That is, I am getting two different equations when considering the upward and downward motion of the ball (given that the upward direction is positive). How do I come up with a formula for $$v(t)$$ in such a scenario?

For context, I am self studying classical mechanics from the book - 'Classical Mechanics' by John R. Taylor.

For differential equations, you should avoid thinking in terms of vector norms. The ODE in terms of vectors is: $$m\frac{d\vec v}{dt} = m\vec g+\vec f\\ \vec f = -\alpha \vec v$$ with $$\alpha>0$$ the linear friction coefficient. Projecting in a direction, say $$z$$ you get: $$m\dot v_z = mg_z-\alpha v_z\\$$ The equation is the same no matter the direction of projection, whether it is $$+e_z$$ or $$-e_z$$ which differ by a global sign in the equation which is canceled out.
If you really want to think in terms of norm, then you need to assume that all three vectors are collinear (true iff it is the case in the initial condition). You made a sign mistake in your analysis. When the ball is going up, gravity points down and friction as well so: $$m\dot v = -mg-\alpha v$$ When the ball is going down, so is gravity, but friction is pointing up: $$m\dot v = mg-\alpha v$$ This is consistent with the previous projection analysis. Btw, you know the relative sign of acceleration and friction force since it must damp the motion.