4 levels of projectile motion

2 January 2026

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Abstract

Inspired by the cooking videos of the format "n levels of [dish]", I’m here applying a similar concept to scientific modeling. I will be comparing 4 levels of projectile motion, from the easiest "textbook" model, which is the simplest of all and assumes only constant gravity in a flat Earth, to a more complex model taking into account factors like quadratic drag, buoyancy, and the Magnus effect.

Level 1: just constant gravity in a vacuum

This is the easiest level of all. It is ubiquitous in all middle and high school level textbooks on physics (Serway & Jewett, 2016), and it’s commonly used as a way to teach the basic idea of 2D kinematics, and tracking the position as a function of time of an object in the ℝ2 plane using parametric equations of the form x = f(t), y = g(t). Considering gravity in the negative y direction to be the only force acting on the object, and, as Galileo noticed, if gravity is the only force on an object, the mass is not involved in the motion. We assume the object is significantly close to Earth’s surface, in which case constant gravity in a single direction can be assumed. With all of this information, Newton’s second law gives the simple equations $$\begin{align} \ddot{x}=0, \\ \ddot{y}=-g \end{align}$$ where g = 9.8 m/s2 is the standard Earth surface gravity. These equations are the "trivial" kind of ODE, in which simply integrating both sides twice suffices to solve them. Assuming the particle begins with initial speed u at the origin, the solutions are simply $$\begin{align} x(t) = ut \cos \theta, \\ y(t) = ut\sin \theta - \frac{1}{2}gt^2, \end{align}$$ where θ is the angle of launch with respect to the horizontal. Eliminating t, the trajectory can be easily found to be $$y=x\tan(\theta)-\frac{gx^2}{2u^2\cos^2(\theta)},$$ which is a parabola, with an inverse "U" shape in the xy plane. From this the range, the area swept and more can be computed. This case won’t be discussed further here, as the reader is most likely familiar with the dynamics, which they almost certainly covered in their secondary school education. This case is covered with detail in introductory books like (Serway & Jewett, 2016) (Knight, 2008).

Level 2: constant gravity with a linear drag force

Finding a theoretical formula for the drag force is a well-known problem in fluid mechanics, that notoriously requires heavy usage of experimental and numerical rather than pure analytic methods. However, for low Reynolds numbers (Re < 1), there exists a very simple solution for the drag as a function of the speed v (Acheson, 2005). It is known as Stokes’ law, and it is a relation of the form F = −bv, where, e.g., in the case of a sphere of radius r, b = 6πηr, where η is the viscosity of the medium. This is the most complex case that admits a closed form analytical solution (Taylor, 2005). The net force acting on the object in this case is F = −mgĵ − bv. So, resolving the resulting equation of motion into the x and y components gives $$\begin{align} m\ddot{x}=-b\dot{x} \\ m\ddot{y}=-mg-b\dot{y} \end{align}$$ These equations can be solved through separation of variables, or via the general theory of linear constant coefficient 2nd order differential equations (Logan, 2015). Either way, the solutions are found to be $$\begin{align} x(t)=\frac{mv_x(0)}{b}(1-e^{-\frac{bt}{m}}) \\ y(t)=\bigg( \frac{mg}{b} + v_y(0))\bigg )\frac{m}{b}(1-e^{-\frac{bt}{m}})-\frac{mgt}{b}. \end{align}$$ These solutions imply that the projectile travels through a parabola-like curve that becomes steeper as the projectile gets closer to the ground. A trajectory equation similar to (3) can be found by eliminating t from equations (5), or alternatively, via a differential equation (Borghi, 2013): $$y=\frac{v_y(0)+v_{\mathrm{T}}}{v_x(0)}x+\frac{m^2g}{b^2}\ln\bigg( 1-\frac{mx}{bv_x(0)} \bigg),$$ where $v_\mathrm{T}=\frac{mg}{b}$. In this case, there is no general closed form analytical solution for the drag and it must be computed either numerically via e.g., Newton’s method, or via perturbations (Taylor, 2005). This kind of equations should only be used in the physical world to model e.g., the motion of a microscopic object like a bactetria. As with the previous level, this case won’t be treated much further here, since it is a standard problem in undergraduate level textbooks (Taylor, 2005) (Greiner, 2004)

Level 3: constant gravity with a quadratic drag force

We now begin to examine the most realistic models. We’re now entering into terrain where there’s no general closed form analytical solution, though (Parker, 1977) gives some approximate analytical solutions for both the the short and long time regimes. I will present some Python code with the numerical integrations. In this case, the force acting on the projectile is almost the same in the previous section, but with the drag force instead being Fdrag = −c|v|v. In vector form, thus, the equations take the form $$m\ddot{\mathbf{r}}=-mg\hat{j}-c|\dot{\mathbf{r}}|\dot{\mathbf{r}}.$$ Which in terms of components, turns into $$\begin{align} m\ddot{x}=-c\dot{x}\sqrt{\dot{x}^2+\dot{y}^2} \\ m\ddot{y}=-mg-c\dot{y}\sqrt{\dot{x}^2+\dot{y}^2}. \end{align}$$ These equations are now coupled and non-linear, making a general closed form analytical solution impossible. Highly accurate approximations for the maximum height, arrival time, and flight range are given in (Turkyilmazoglu, 2016). As per the so-called "drag equation", the coefficient c can be found to be $c=\frac{1}{2}C_d A\rho$, where Cd is the so-called "drag coefficient", A is the cross-sectional area and ρ is the density of the medium. To visualize this case, it’s better to proceed with an example. Here, we’ll procede with a horizontal throw case application of equations (8). We will model a cannon ball with density ρb = 707 kg/m3, corresponding to the density of certain kinds of stone, and a radius of 18.25 cm. Here’s the plotted trajectory of a sample numerical integration of equations with time step h = 0.001 (8) in Python with Heun’s method, a second order method (Süli, 2003):
image
This image shows the path equations (8) predict for a particle initially at rest along the vertical direction, starting from x(0) = 0, y(0) = 100 m and u = 200 m/s. The image is then compared with what would the path been like without drag. One of the things we can notice from the image is that the path of the particle with drag is steeper than the path without drag. Also, that the range has been significantly reduced, the particle without drag goes farther than the one with drag. This happens because the drag reduces the velocity of the particle as time passes. It can be seen from equations (8) that if the velocity is always positive, the acceleration will always be positive, meaning the velocity is always decreasing in both the x and y directions. The model presented in this section is good for Reynolds numbers Re > 1000.

Level 4: Constant gravity with quadratic drag, buoyancy, Magnus force

This level is the most realistic out of the ones presented so far. It still models the behavior with a deterministic differential equation, but takes the Magnus force into account. Aditionally, this model is now 3D. Applications of this model include sports. For example, it’s useful to model the dynamics of a ball after a footballer kicks a penalty, both the trajectory and where it ends up landing, as a result of simply the initial kick the footballer gives. Making it therefore useful, e.g., in game development to make FIFA/PES-like games. The mass here is treated like a rigid body. Therefore, these equations, properly speaking, track the trajectory of the center-of-mass of the ball. The Magnus force acting on a given object can be modeled by a simple expression: FMagnus = Sω × v (Fitzpatrick, 2004). Experiments to determine the value of the Magnus force coefficient expression are given in (Goff, 2013). We can also add buoyancy here. It also has a simple expression, the buoyant force is the removed mass times the gravitational acceleration: Fb = −ρfVgk̂. Combining all of this gives the equation: $$m\ddot{\mathbf{r}}=-mg\hat{k}+\rho_b Vg\hat{k}-c|\dot{\mathbf{r}}|\dot{\mathbf{r}}+S\mathbf{\omega}\times\mathbf{\dot{\mathbf{r}}}.$$ Like in the previous section, we can divide this into components, though this time there are 3 of them: $$\begin{align} m\ddot{x}=-c\dot{x}\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}-S\omega\dot{y} \\ m\ddot{y}=-c\dot{y}\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}+S\omega\dot{x} \\ m\ddot{z}=g\bigg(\frac{\rho_f V}{m}-1\bigg)-c\dot{z}\sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} \end{align}$$ Like in the previous case, an analytical solution is impossible. Though it’s possible to derive asymptotic special cases from this equation, as done in (Turkyilmazoglu, 2020). This is what a python script solving equations (10) numerically with initial conditions x(0) = y(0) = z(0) = 0 and ẋ(0) = 7 m/s, ẏ(0) = 5 m/s, ż(0) = 10 m/s, for a football with a mass of 450 g, and a radius of 11.15 cm:
image
This shows how the trajectory is deviated horizontally thanks to the Magnus force, as well as of course the steeper trajectory derived from the drag force. In this case, the numerical integration took the time step h = 0.01, and took a total of 10000 steps to produce this figure. It’s interesting to note how the Magnus force has a similar form to the magnetic force F = qv × B, making this Magnus force mimick the effect a magnetic field has over a charged particle.

Discussion

In this pedagogical paper, we have revised 4 different levels of complexity for the projectile motion physical problem. We have increased our complexity from the easiest, textbook model that is simple but unrealistic in many cases to the most complex model that describes more things accurately, e.g., a football’s trajectory after a penalty is taken. This paper shows how can complex phenomena be modeled, and how models take into account more and more variables as the physical world demands it.

References