Solve Equation of Motion when gravity is two dimentional

Question

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)

Answer 1

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.

Answer 2

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 + S\cos(\theta) t + \tfrac12 g_x t^2, \tag1\\ d_y &= y + S\sin(\theta) t + \tfrac12 g_y t^2, \tag2 \end{align}

let $\mathbf x = (x,y),$ $\mathbf d = (d_x,d_y),$ $\mathbf v = (S\cos(\theta),S\sin(\theta)),$ 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 $\lVert\mathbf v\rVert = S.$ 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): $$ \lVert\mathbf d - \mathbf x\rVert^2 - t^2(\mathbf d - \mathbf x)\cdot \mathbf g + \tfrac14 t^4 \lVert\mathbf g\rVert^2 = t^2 \lVert\mathbf v\rVert^2 = t^2 S^2. $$ Collect all terms on one side and rearrange them to get $$ \tfrac14 t^4 \lVert\mathbf g\rVert^2 - t^2((\mathbf d - \mathbf x)\cdot \mathbf g + S^2) + \lVert\mathbf d - \mathbf x\rVert^2 = 0. \tag3 $$ If we set $a = \tfrac14 \lVert\mathbf g\rVert^2,$ $b = -((\mathbf d - \mathbf x)\cdot \mathbf g + S^2),$ $c = \lVert\mathbf d - \mathbf x\rVert^2,$ 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, $$ t^2 = \frac{(\mathbf d - \mathbf x)\cdot \mathbf g + S^2 \pm \sqrt{((\mathbf d - \mathbf x)\cdot \mathbf g + S^2)^2 - \lVert\mathbf g\rVert^2 \lVert\mathbf d - \mathbf x\rVert^2}} {\tfrac12 \lVert\mathbf g\rVert^2}. \tag4 $$

Observe that there solutions only if $((\mathbf d - \mathbf x)\cdot \mathbf g + S^2)^2 \geq \lVert\mathbf g\rVert^2 \lVert\mathbf d - \mathbf x\rVert^2 \geq 0$ and if the entire right-hand side of Equation $(4)$ is non-negative (since we must have $t^2 \geq 0$). Also note that $$\sqrt{((\mathbf d - \mathbf x)\cdot \mathbf g + S^2)^2 - \lVert\mathbf g\rVert^2 \lVert\mathbf d - \mathbf x\rVert^2} < \lvert (\mathbf d - \mathbf x)\cdot \mathbf g + S^2 \rvert,$$ 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 $(\mathbf d - \mathbf x)\cdot \mathbf g + S^2 \geq \lVert\mathbf g\rVert \lVert\mathbf d - \mathbf x\rVert \geq 0$ then there is at least one possible value of $t^2,$ and if $(\mathbf d - \mathbf x)\cdot \mathbf g + S^2 > \lVert\mathbf g\rVert \lVert\mathbf d - \mathbf x\rVert$ 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 $t^2.$