# 3D ballistic trajectory with quadratic drag. Calculating position and velocity at time $t$

A particle starts at the origin and has an initial velocity represented by a 3D vector. The particle experiences gravity and air resistance with quadratic drag (based on velocity^2). What I've been looking for are parametric equations for $x$, $y$, and $z$ position and velocity of the particle at a given time $t$.

To keep answers similar the two constants will be g as gravity and a as the air resistance. Also assume $z$ is positive elevation. $v_x(t)$ will be the velocity for the $x$ component at time $t$. ${v_x}_0$ would represent the initial velocity for the $x$ component.

At the bottom of this wikipedia article it suggests there's an analytic solution to the 2D problem by a teenager Shouryya Ray. The corresponding Math.SE and Phys.SE links have people discussing solving for a constant of friction, but there's no parametric equations given anywhere. I'm not sure how to go from what they're talking about to the actual parametric equations I need.

To cover what I've learned so far. In 3D with gravity and no drag the following parametric equations can be used:

$v_x(t) = {v_x}_0$

$v_y(t) = {v_y}_0$

$v_z(t) = {v_z}_0-gt$

$s_x(t) = {v_x}_0t$

$s_y(t) = {v_y}_0t$

$s_z(t) = {v_z}_0t - \frac{1}{2} g * t^2$

With linear drag per the Wikipedia article we'd used:

$v_x(t) = {v_x}_0e^{-\frac{a}{m}t}$

$v_y(t) = {v_y}_0e^{-\frac{a}{m}t}$

$v_z(t) = -\frac{mg}{a}+({v_z}_0+\frac{mg}{a})e^{-\frac{a}{m}t}$

$s_x(t) = \frac{m}{a}{v_x}_0(1-e^{-\frac{a}{m}t})$

$s_y(t) = \frac{m}{a}{v_y}_0(1-e^{-\frac{a}{m}t})$

$s_z(t) = -\frac{m * g}{a}t + \frac{m}{a}({v_z}_0 + \frac{m*g}{a})(1 - e^{-\frac{a}{m}t})$

With quadratic drag we have the velocity's length so each component's velocity and position over time depends on the other velocity components. Starting with this post I can write the 3D version for the first part:

$v_x' = -av_x*\sqrt{v_x^2 + v_y^2 + v_z^2}$

$v_y' = -av_y*\sqrt{v_x^2 + v_y^2 + v_z^2}$

$v_z' = -av_z*\sqrt{v_x^2 + v_y^2 + v_z^2}-g$

But I lack the mathematical background to continue. It's not even clear to me what the constant of friction they calculate is used for since it's not based on time. I would think any trick that attempts to calculate a constant coefficient would need to know about t. I'm probably thinking of this wrong. Any help would be appreciated as I imagine this is a common problem in physics.

edit: Also my problem case is specifically for a constant gravity vector if that helps for specific solutions so the path must be a parabola of sorts. As far as I can tell there should be a function as there's one y value for every x.

There's no need to work in 3D as the motion is confined to a 2D plane. Just choose your $x$ and $y$ axes to lie in that plane.