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I have a projectile in 2D space $r=(x, y)$ at time $t_0=0$, which has an initial velocity $v_0$, a launch angle $\theta$ from $(1, 0)$ and which accelerates with a constant $a_0$ until time $t_1$ in the current flight direction of the projectile as well as a constant $g$ downwards $(0, -1)$.

This is, for example, a simplified model of a rocket with a short-lived motor, ignoring changes in mass from the propellant and any air drag.

I'm looking for a definition of the flight trajectory, so that I can determine functions describing the angle $\theta$ to hit a point $(x, y)$, the time to get there, and similar. I only found https://cnx.org/contents/--TzKjCB@8/Projectile-motion-on-an-incline so far. I planned to use the given formulas there to piece together a case distinction based on whether the time to target is smaller or larger than $t_1$, but I am not really sure how to connect the "ends" of the two cases and with the acceleration vector changing over time, I don't know if this can even be done in this way.

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  • $\begingroup$ What do you want to do if you have an answer? Is it a school exercise? Do you want to send a Rocket to Russia? Or what? $\endgroup$ – Deschele Schilder Sep 2 '20 at 20:16
  • $\begingroup$ @descheleschilder No it's for an external tool (program) to a game, i.e. a hobby project. $\endgroup$ – dama Sep 2 '20 at 21:02
  • $\begingroup$ Alright then! I'm curious if you'll get it finished, $\endgroup$ – Deschele Schilder Sep 2 '20 at 21:49
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The flight path is given by: $$\frac{d^2}{dt^2}\vec r(t) = a_0(t) \frac{d}{dt}\hat r (t)+ \vec g$$ where $$a_0(t)=\begin{cases} a_0 & t<t_1 \\ 0 & t_1<t \end{cases}$$ and $$\hat r (t) = \frac{\vec r (t)}{||\vec r (t)||}$$

I put this differential equation into Mathematica, as well as a simpler differential equation involving just the initial portion where $a_0(t)=a_0=const.$. In both cases Mathematica was unable to evaluate it using DSolve, so unfortunately, this does not appear to have an analytical solution. It will need to be solved numerically, which I did using NDSolve.

For a fairly brief rocket burn, this produces a trajectory that is nearly parabolic:

Nearly parabolic trajectory

Perhaps more interesting is a trajectory with a long rocket burn. This trajectory it seems that the rocket gradually tips over and then propels itself into the ground on a decidedly non-parabolic trajectory:

Strongly non-parabolic trajectory

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  • $\begingroup$ Would it have an analytical solution if $a(t)$ did not change over time i.e. for the subproblem $ t < t_0$? $\endgroup$ – dama Sep 3 '20 at 10:57
  • $\begingroup$ @dama No, I tried that too. That is precisely the part that makes the problem non-analytical $\endgroup$ – Dale Sep 3 '20 at 11:06
  • $\begingroup$ I accepted your question since it nicely answers the physics side of the question, but considering the "openness" of differential equations it would be nice to know what you tried or where and why you stopped. The best i have atm is that this boils down to a "system of coupled, non-linear differential equations" and that topic is kind of broad AFAICT. $\endgroup$ – dama Sep 3 '20 at 21:30
  • $\begingroup$ @dama Sure. I updated the answer with that info $\endgroup$ – Dale Sep 4 '20 at 0:06
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At $t=t_0=0$, the object starts to accelerate an angle $\theta _0$ with the $x$-axis. The acceleration is in the minus $y$-direction with a magnitude of $1$. As you stated.
How changes the velocity of $v_0$ when the object is launched? The x-component of $v_0$ is equal to $v_{0,x}=v_0 \cos {(\theta _0)}$.
The y-component is $v_{0,y}=v_0 \sin (\theta _0)$ . How high will the object rise? Well, $h=\frac 1 2 (a_0\sin{\theta _0}-g)^2 t_1^2=\frac 1 2{t_1^2(a_0\sin{\theta _0}-g)}^2$.

So the time it takes to reach $y=h$ will be $t_1=\sqrt{\frac {2h} {a_0\sin{\theta _0}-g}}$. At the time $t_1$ the vertical velocity will be $v_v=v_0-(a_0\sin{\theta _0}-g)t_1$.
The distance traveled in the x-direction (until the acceleration stops) is $x=\frac 1 2 a_0 {t_1}^2\cos{\theta _0}$. The velocity in the x-direction will be $v_x=a_0\cos{\theta _0}t_1$

So, now we know the x- and y-velocities, as well as the distances x and y, traveled until the engine stops, we can calculate the second part of the trip. How? Assuming the object still has a component upwards, the last part will be a part of a parabola (as was the first part). The initial x- and y-velocities are known, as well as the initial x and y value for the second free fall part.

Put them together and there you go. I'll leave that for you to calculate. I gave you the basic recipe and ingredients.

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  • $\begingroup$ I don't see where you model that acceleration at any given time is in the current direction of flight. i.e. the angle at which $a_0$ applies is not constant . $\endgroup$ – dama Sep 1 '20 at 20:15
  • $\begingroup$ I let the object accelerate from the initial position and velocity to reach the point where only $g$ acts upon the object. True, the angle varies with time, but that's been taken into account implicitly. $\endgroup$ – Deschele Schilder Sep 1 '20 at 20:18
  • $\begingroup$ I edited the answer for you. $\endgroup$ – Deschele Schilder Sep 1 '20 at 20:23

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