Vertical trajectory of a projectile with quadratic drag

Question

I was wondering is there is an equation (or function) that can predict the apogee time and height of an object given its current acceleration and the usual constants ( mass, drag coefficient, cross area, etc.)

I have recently taken AP physics Mech, so I have some experience solving these types of problems but could not solve this one. I have tried using energy, momentum, even modified kinematics. All of which ended in dependency loops. Drag affects velocity and velocity affects drag.

The drag needs to be quadratic and it would be nice if there was a way to incorporate how air density changes throughout the flight path since that follows a function (given the current altitude) and that affects drag. Thanks for your help!

Recent Update

I had recently come in contact with a cousin of mine that redirected me to this formula which basically solves my question except for two things:

https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/flight-equations-with-drag/

1. It needs initial velocity rather than initial acceleration
2. The drag equation (masked within $V_t^2 $ in the final $y_{\text{max}}$) does not account for a changing air density as altitude increases (I am given initial altitude). Though I am no longer entirely sure this is needed because the max height of the projectile is about 800ft.

Answer 1

Yes, if your trajectory is vertical, there is a simple analytic solution for a purely quadratic drag. In dimensionless units (unit mass, terminal velocity speed unit), you need to solve: $$ \dot v + v^2\text{sgn}v+1 = 0 $$ You can solve it by separation of variables. It is best to divide the solution prior to the apogee and after the apogee, which I will set at $t=0$. Prior to the apogee, $v>0$: $$ \dot v + v^2+1 = 0 $$ so the apogee time starting from speed $v_0$ is: $$ t_0 = \arctan v_0 $$ The full solution is (origin of the vertical coordinate being the apogee): $$ v = -\tan t \\ z = \ln(\cos t) $$ Starting from upward speed $v_0$, the apogee height is: $$ h = \frac12\ln(1+v_0^2) $$ You'll notice that the solution has a finite time blowup in the past at $t=-\pi/2$. This is due to the quadratic divergence of the drag for infinite positive velocity, which is faster than linear.

You can also integrate the equation after the apogee where $t>0,v<0$: $$ \dot v-v^2+1 = 0 $$ You get: $$ v = -\tanh t \\ z = -\ln(\cosh t) $$ so that the velocity does relax exponentially to the terminal velocity.

For completeness, if your point started at a downward velocity faster than the terminal velocity, there is no apogee and it will just slow down to the terminal velocity. The solution is: $$ v = -\coth(t-t_b) $$ with $t_b$ the past blowup time, again due to the quadratic divergence of the drag for negative velocities.

From there, you can extend these analytic results for almost vertical trajectories by perturbative methods. For a general solution, you'll need numerical methods.

Answer 2

(Elaborating on @LPZ's [solution 283846]) If you solve for arbitrary constant of drag $d$, and gravitational acceleration $a$, using $$\dot v=-a-d\ \text{sgn}(v)\,v^2$$ this leads to a concise expression for the shape of the trajectory: $$ h(t)=- \frac{\theta\, (-t)}{d}\ \log\,\cos (\sqrt{ad}\, t) - \frac{\theta\, (t)}{d}\ \log\,\cosh (\sqrt{ad}\, t) $$ where $\theta$ is the step function [see Wikipedia], defined by $\theta(t)=0$ for $t<0,$ and $\theta(t)=1$ for $t>0$. The solution can of course be shifted in time and height by $$ h(t) \rightarrow h(t-t_\text{max})+h_\text{max}, $$ but that doesn't alter the shape.

And the parameter $a$ will just stretch the shape by $1/\sqrt a$ horizontally. Similarly, the parameter $d$ will only stretch the shape by $1/d$ in the vertical direction and by $1/\sqrt d$ horizontally. When plotted for $a=9.8$ and the cases $d\rightarrow 0, d=1,$ and $d=2$, it looks like this:

The blue line is the parabola for $d\rightarrow 0$ with the well-known drop of $4.9$m in the first second after maximum height