Csdrhrt

How does one solve the following system of equation for Θ. Only unknown variables are Θ and t.

This is the equation of motion when gravity is two dimensional. WolfarmAlpha succeeded to solve but I fail to understand the solution.

$d_0$ is $d_x$ and $d_1$ is $d_y$,

$g_0$ is $g_x$ and $g_1$ is $g_y$

(WolframAlpha only allows single letter variable names)

This is still gravity in 1D. Just solve the problem with one axis along the vector sum of $g_0$ and $g_1$, and the other axis perpendicular to it.

Based on some of its answers, I suspect Wolfram Alpha does not always treat the same variables as "known" that you do.

Starting with these equations,
begin{align}
d_x &= x + Scos(theta) t + tfrac12 g_x t^2, tag1\
d_y &= y + Ssin(theta) t + tfrac12 g_y t^2, tag2
end{align}

let $mathbf x = (x,y),$ $mathbf d = (d_x,d_y),$
$mathbf v = (Scos(theta),Ssin(theta)),$ and $mathbf g = (g_x,g_y).$
then Equations $(1)$ and $(2)$ can be expressed by a single vector equation:
$$
mathbf d = mathbf x + t mathbf v + tfrac12 t^2 mathbf g ,
$$
where $mathbf v$ is an unknown vector such that $lVertmathbf vrVert = S.$
Collect everything except the $t mathbf v$ term on one side:
$$
mathbf d - mathbf x - tfrac12 t^2 mathbf g = t mathbf v.
$$
Square both sides (that is, take dot product of the vector with itself, or in other words, compute the square of the magnitude):
$$
lVertmathbf d - mathbf xrVert^2 - t^2(mathbf d - mathbf x)cdot mathbf g
+ tfrac14 t^4 lVertmathbf grVert^2 = t^2 lVertmathbf vrVert^2 = t^2 S^2.
$$
Collect all terms on one side and rearrange them to get
$$
tfrac14 t^4 lVertmathbf grVert^2
- t^2((mathbf d - mathbf x)cdot mathbf g + S^2)
+ lVertmathbf d - mathbf xrVert^2 = 0. tag3
$$
If we set $a = tfrac14 lVertmathbf grVert^2,$
$b = -((mathbf d - mathbf x)cdot mathbf g + S^2),$
$c = lVertmathbf d - mathbf xrVert^2,$ and $u = t^2,$
then Equation $(3)$ becomes
$$ au^2 + bu + c = 0,$$
which is a quadratic equation in $u$ and (if it has any solutions at all)
has solutions only of the form
$$ u = frac{-b pm sqrt{b^2 - 4ac}}{2a}.$$

That is,
$$
t^2 = frac{(mathbf d - mathbf x)cdot mathbf g + S^2 pm
sqrt{((mathbf d - mathbf x)cdot mathbf g + S^2)^2
- lVertmathbf grVert^2 lVertmathbf d - mathbf xrVert^2}}
{tfrac12 lVertmathbf grVert^2}. tag4
$$

Observe that there solutions only if
$((mathbf d - mathbf x)cdot mathbf g + S^2)^2 geq
lVertmathbf grVert^2 lVertmathbf d - mathbf xrVert^2 geq 0$
and if the entire right-hand side of Equation $(4)$ is non-negative (since we must have $t^2 geq 0$).
Also note that
$$sqrt{((mathbf d - mathbf x)cdot mathbf g + S^2)^2
- lVertmathbf grVert^2 lVertmathbf d - mathbf xrVert^2}
< lvert (mathbf d - mathbf x)cdot mathbf g + S^2 rvert,$$
which rules out the possibility that
$(mathbf d - mathbf x)cdot mathbf g + S^2 < 0$
(because if that inequality were true, the right-hand side of Equation $(4)$ would be negative).
But if
$(mathbf d - mathbf x)cdot mathbf g + S^2 geq
lVertmathbf grVert lVertmathbf d - mathbf xrVert geq 0$
then there is at least one possible value of $t^2,$ and if
$(mathbf d - mathbf x)cdot mathbf g + S^2 >
lVertmathbf grVert lVertmathbf d - mathbf xrVert$
there are two possible values.

Two possible values of $t^2$ means there is a "high" trajectory and a "low" trajectory, both of which pass through the given target point.
Of course, for each value of $t^2$ there are two possible values of $t,$
one positive and one negative.
A positive value of $t$ corresponds to a projectile
that leaves $mathbf x$ at speed $S$ at time $0$ and later arrives at $mathbf d$
(at time $t$),
whereas a negative value of $t$ corresponds to a projectile that first passed through $mathbf d$ at time $t$ (before time $0$) and then arrived at $mathbf x$ at speed $S$ at time $0.$
There are always these two possibilities (if the problem has a solution and $t neq 0$) because trajectories of this kind are reversible.
Assuming you want only non-negative values of $t,$ however, you can take the square root of $t^2.$