How can you model how far a spinning cylinder will move forward during free fall? I am a high school student and have just started calculus. Part of my project involves doing this experiment where I drop a cylinder from about 10m high with a large backspin and see how far forwards it moves. Consider this diagram:

The Magnus force acts perpendicular to the velocity of the cylinder in the fluid and acts in the x-direction because of the pressure difference between the sides of the cylinder. Furthermore, as time goes and the cylinder free fall its velocity in the fluid changes, hence the direction of the Magnus force is not constant and hence changes during the free fall.


My problem is that I have limited knowledge of calculus and how to deal with these changing variables. I have done preliminary experiments and notice a displacement of about a metre at roughly 2000 rpm with a quiet heavy cylinder.
Any help in knowing where to start in terms of modelling this would be extremely helpful and greatly appreciated  :) as I don't want to settle with just having to explain it in layman's terms, it doesn't feel like I fully am able to describe the phenomenon then.
 A: Update: I was so interested answering this question that I made a little simulation of the effect. See: https://joeiddon.github.io/magnus_effect/.
Having skimmed the Wikipedia page on the Magnus Effect, we can take it that the Magnus force, $F_m$, is proportional to the velocity of the cylinder, and perpendicular to it. Note that in reality it depends on other factors, like the cylinder's angular velocity and radius, as well as the density of the fluid (here air), but I will leave this for you to add in, we will just take it to be proportional to velocity.
$$F_m \propto v = kv \label{1}\tag{1}$$
We also know that there is the downwards force due to gravity. This is proportional to the mass of the cylinder, but over a small distance can otherwise be taken to be constant. Ignoring other aerodynamic effects, these are the only two forces on the cylinder: $F_m$, the curving magnus force that acts perpendicular to the cylinder's velocity, and $W$, the weight which acts downwards.
To solve this problem, I can see two approaches: either approximate the change in angle to be zero, so that the Magnus force is always to the left and the weight is the only vertical force, or we can try and form the equation of motion for the cylinder and solve exactly.
I will give the latter approach a go first.
Let the velocity of the cylinder have two components: horizontal $v_x$, and vertical $v_y$. We now try and work out the forces on the cylinder in terms of these, i.e. split up in terms of there vertical components.
Let's start without the Magnus effect. If the only force on the cylinder was its weight then the acceleration in the x-direction would be zero, $\dot{v_x} = 0$, and the acceleration in the y-direction would be minus the acceleration due to gravity, $\dot{v_y} = -g$.
These would be our equations of motion, and we could go ahead and solve.
Now let's try and add in the Magnus effect. Because this force is perpendicular to the velocity and proportional to it (Equation \ref{1}), its x-component will be proportional to the y-component of the velocity, and its y-component will be proportional to the x-component of the velocity. This may not be intuitive to you, personally I drew a diagram with an angle $\theta$ (as you started to in your question), and went through the geometry of the situation before concluding that the x-component of the force was $v\cos\theta$, which is of course the y-component of the velocity, and similarly the y-component of the force was $v\sin\theta$ which is the x-component of the velocity. If you want to understand this more, take a look at circular motion which is the fundamental basis of this idea.
So, algebraically, we can write the Magnus force as a vector (writing as a vector because can't subscript $F_m$!):
$$\vec F_m = \pmatrix{kv_y \\ kv_x}$$
Now we can say that the total force on the cylinder, $F$, is given by:
$$\vec F = \pmatrix{kv_y \\ kv_x - W}$$
which is just adding in the vertical force due to gravity.
We can now use Newton's second law, $\vec F = m\vec a$ to find out equations of motion.
$$\vec a = \frac{1}{m}{\vec F} = \frac{1}{m}\pmatrix{kv_y \\ kv_x - W}$$
Which means our equations are:
\begin{align}
\dot{v_x} &= \frac{k}{m}v_y \\
\dot{v_y} &= \frac{k}{m}v_x - \frac{1}{m}W
\end{align}
This system of differential equations can be solved to find the velocity as a function of time. Integrating the vertical velocity will allow you to find the time for the cylinder to reach the ground since you know the initial height, and then you can integrate the horizontal velocity up to this time to find the distance travelled in the x-direction!
I will not solve this for you, but this is a well defined calculus problem that you can easily research the method of solving (the Mathematicians have the easy job, the Physics is the hard bit!).
Just for closure, I will show you how to solve the easier approximate method I thought of.
Here we say that we can say that the Magnus effect is small in comparison the weight, so that the vertical velocity is unaffected by the Magnus force. This makes our equations of motion much simpler:
\begin{align}
\dot{v_x} &= \frac{k}{m}v_y \\
\dot{v_y} &= \frac{1}{m}W = g
\end{align}
To solve these simpler equations for the horizontal displacement (our goal), we will use the second to find the time to reach the ground, and then use the first to find the horizontal displacement.
The second equation describes a constant vertical acceleration, $a_y = g$, this means there are five general equations called the SUVAT equations that apply to this situation. We can quote these results without having to derive them.
We know the initial vertical height, $s=10m$, and the initial velocity  $u=0ms^{-1}$, and we want the time, $t$, so we choose the equation which involves these variables:
$$s = ut + \frac{1}{2}at^2$$
which gives our time as $t = \frac{2s}{a} = \frac{2s}{g}$.
Now we know the time for the cylinder to reach the ground, we can use the first equation of motion to find the horizontal displacement. We see that this equation involves $v_y$ which is the vertical velocity as a function of time. Luckily the SUVAT equations provide us with this as we have constant acceleration in the y-direction: $v_y = u + at = gt$. Putting this into the first equation gives
$$\dot{v_x} = \frac{k}{m}gt.$$
So the horizontal acceleration is changing with time. To find the horizontal displacement at time $t=\frac{2s}{g}$, we must integrate this equation, twice!
First we integrate to find the horizontal velocity as a function of time.
$$v_x = \frac{k}{2m}gt^2 + c$$
And the horizontal velocity is initially zero, so $v_x(t=0) = 0$, so $c=0$.
Now we can once again integrate our velocity to find the horizontal displacement as a function of time.
$$s_x = \frac{k}{6m}gt^3 + c'$$
Once again the initial horizontal displacement is zero, so $c' = 0$.
Therefore we finally get to our end equation for horizontal displacement (with approximation of course), which is:
$$s_x = \frac{k}{6m}gt^3$$
Plugging in our time of $\frac{2s}{g}$ gives the final horizontal displacement when the cylinder hits the ground of $\frac{4sk}{3mg^2}$.
There are definitely mistakes in these calculations (the final result isn't even dimensionally consistent, so comment the fixes please!) but they should definitely set you on your way - I am confident of the correctness of the equations of motion, and the process of forming them should help.
