I have a body $\vec{b}$ in a frictionless, gravityless plane, with an initial velocity $\vec{v}_i$.
I want to redirect this body so that it's final velocity $\vec{v}_f$ has the same magnitude as $\vec{v}_i$, but towards an arbitrary point $\vec{t}$ anywhere on the plane after the redirection.
This redirection is done by applying an acceleration $\vec{a}$, constant both in magnitude and in direction, such that the redirection trajectory is parabolic, on $r$, until the velocity is aligned with $\vec{t}$.
Knowning:
- The initial position of the body $\vec{b}$, on $r$
- The initial velocity of the body $\vec{v}_i$, tangent to $r$ at $\vec{b}$
- The final velocity magnitude is the same as the initial velocity: $||\vec{v}_i||=||\vec{v}_f||$. It also implies that the end of the redirection $\vec{e}$ is the symetry of $\vec{b}$ on $r$
- The magnitude of the acceleration $||\vec{a}||$
- The final velocity $\vec{v}_f$ has the same direction as $\vec{et}$
Problem: Find the parabola $r$: either the direction of $\vec{a}$, the position of $\vec{i}$, or angle $\alpha{}$, knowing $\vec{b}$, $\vec{t}$, $\vec{v}_i$ and $||\vec{a}||$
What I tried: I see an overlap between this problem and projectile motion. The equations there are helpful, since they link the kinematic inputs with the geometric problem. The analogy is is that the angle between the initial velocity and the acceleration is unknown, and the 'range' of the projectile, $||\vec{be}||$, is not given in the problem above, since the 'command' is not the range, but instead that the body 'is heading towards a point underground' when it hits the ground.
Let's define: $$ a=\|\vec{a}\|,\quad v=\|\vec{v}_i\|,\quad \theta={\alpha\over2},\quad d=\|\vec{be}\|,\quad l=\|\vec{bt}\|,\quad \phi=\angle ibt. $$ If $t$ is the time taken to travel from $b$ to $e$, then we have: $$ d=t\cdot v\sin\theta,\quad a={2v\cos\theta\over t}. $$ Eliminating $t$ we thus find $$ ad=v^2\sin2\theta. $$ On the other hand, from the sine rule we have: $$ {l\over\sin\left({\pi\over2}+\theta\right)}= {d\over\sin\left(\pi-\phi-2\theta\right)} \quad\implies\quad d={l\sin\left(\phi+2\theta\right)\over\cos\theta}. $$ We can eliminate $d$ from the last two equations, to find an equation for $\theta$: $$ al\sin\left(\phi+2\theta\right)=v^2\sin2\theta\cos\theta. $$ Unfortunately, this is not easy to solve and some numerical method should be used. Once you have $\theta$ and $d$, focus and directrix of the parabola can be readily found.