# Vertical trajectory of a projectile with quadratic drag

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.
• You're looking for an analytic solution as opposed to a numerical one? If you're trying to do a numerical solution with given initial conditions, I would look into Verlet or Runge-Kutta methods. Commented Jun 21 at 2:31
• What is stopping you from just integrating? It is not exactly simple, but also not tremendously difficult. Instead, it is the mixture of both linear and quadratic drag that is much more difficult. Or when you want to compare linear drag v.s. quadratic drag's times. Commented Jun 21 at 3:06
• It might be beneficial to write out the formulae you are using that lead to the dependency loop. Commented Jun 21 at 11:16
• @Kyle Kanos, one such equation I was using is A = g + (bv^2)/m but v = integral of A, so then there is a definition incongruity. Also mgh = 0.5mv^2 -integral of drag by distance (h in this case). However h is what I am trying to find so there is another loop. Thanks! Commented Jun 30 at 23:37

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.

(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

• Thank you and @LPZ for your responses! I did try really hard to understand this, but I have some questions: I don't quite understand what some of the symbols are (like z, or sgn ) I just need two equations: One that gives out time of apogee and another altitude of apogee. I am given vertical acceleration (drag and gravity) and the constants (mass, drag coefficient, etc.) Also, this is only for the ascent part of the flight path so terminal velocity and post apogee are not needed. Thank you! Commented Jun 24 at 23:43
• You cannot know the trajectory if you only are given the things you mention, you also would need the starting velocity! And the whole discussion, also in @LPZ's answer, assumes that the motion is purely vertical (because your title suggests so, although admittedly it does not explicitly say so...) So can you confirm that you shoot the projectile straight upwards with a given initial speed? Or else you'll never know how high it will get. Commented Jun 29 at 12:22
• I had been given a lead on the answer yesterday and updated the question. I am launching an object in the air in real life so I won't be able to guarantee that it goes in a perfectly vertical path and I only have an accelerometer (it's 3-axis so I can separate the vertical component) and a clock onboard so I am not sure how I could get the initial velocity. Commented Jun 30 at 15:15
• I was wondering if since acceleration and velocity are related, maybe there could be a solution. Or maybe we could back track the acceleration at apogee to our initial acceleration and find how much of a height difference there is? Commented Jun 30 at 15:26
• Perhaps I see what you mean... After shooting it up the air, drag and gravity together give a deceleration, if you know that value, you (indirectly) know how fast it must be going! So OK, then you should not only give the constants $a$ and $d$ but also the value $\ddot v$ directly after launch, to have enough information for determining the trajectory. But the fact that you are not always launching vertically makes it more complicated than what we had until now! Commented Jun 30 at 15:58