Algorithm to construct a parabola that hits a given target and avoids given boundaries
Consider an endpoint $(x_E,y_E)$  of an obstacle line segment.  I'm going to assume for simplicity that if the projectile goes exactly through $(x_E,y_E)$, this is a legal trajectory, i.e., it's OK for a projectile to &quot;just graze&quot; the obstacle.  (If that's not OK, it is easy to adjust the solution below to just miss the corner by a tiny amount $\epsilon$, or it is easy to adjust the size of each obstacle by $\epsilon$ in each direction.)
Let's define a &quot;grazing constraint&quot; as one of the following possibilities:

If the point $E$ is a vertex of an obstacle (i.e., an endpoint of an obstacle line segment) or is $R$ or $S$, then &quot;The parabola must go through $E$&quot; is a grazing constraint.

If $EF$ is an obstacle line segment, then &quot;The parabola intersects $EF$ and is tangent to it&quot; is a grazing constraint.


In this way, given $n$ obstacle line segments, we obtain $2n+2$ possible grazing constraints.
Now I make two claims:
Claim 1: Any two grazing constraints uniquely determine the parabola.
Proof: If both grazing constraints are of the &quot;must go through E&quot; type, then together with the target $T$, we have three points, and any three points uniquely define a parabola, i.e., given three points, there is a unique parabola that goes through those three points.  A similar argument can be made if one or both of the grazing constraints are of the &quot;tangent&quot; type.
Claim 2: If there is any solution to your problem, there is a solution that satisfies two of the possible grazing constraints listed above.
(I don't have a proof for this claim.)
This immediately suggests a simple algorithm: enumerate over all pairs of grazing constraints, find the unique parabola that satisfies those two grazing constraints and goes through the target $T$, and check whether this parabola avoids all obstacles and intersects the starting line segment RS.  You can check whether it avoids all obstacles by iterating over all obstacle line segments, and for each, finding the intersection using the method you described.  If there is a valid solution to your problem, this method will find it.
The running time of this method will be something like $O(n^3)$, where $n$ is the number of obstacle line segments.  I don't know if there is a more efficient method.  There might well be.