Projectile motion on incline with time-limited, constant acceleration 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.
 A: 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:

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:

A: 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.
