physicsprojectile motion
Given:
I want to find the range of a projectile (ignoring wind resistance)
Hours of searching have given no useful results. Those that I thought were useful turned out to be incapable of accounting for negative angles. This formula in particular returns a range of 0 if the angle is 0, which simply doesn't make sense.
For example:
I would like to modify your original equations slightly by affixing subscripts to $v$ and $\theta$ to acknowledge the fact that they use the angle and speed of the projectile's motion at a particular point in space and time, namely, at the initial point $(x,y) = (x_0,y_0)$ and time $t = t_0$, and also to take into account that $x_0$, $y_0$, and $t_0$ are not necessarily all zero. (In particular, in your example in the question, $x_0 = -3$.) The result is \begin{align} x &= x_0 + v_0 (t-t_0) \cos \theta_0, \\ y &= y_0 + v_0 (t-t_0) \sin \theta_0 - \tfrac12 g (t-t_0)^2. \end{align} There are other ways to express this, but this form of the equations is relatively simple to write. (The equation measures the "extra" amount of $x$, $y$, and $t$ past the point $(x_0,y_0)$ and time $t_0$.)
We now consider another point in space and time further along the initial trajectory, at the point $(x,y) = (x_1,y_1)$ and time $t = t_1$. Note that you can uniquely determine $x_1$ and $y_1$ for any $t_1$, you can uniquely determine $t_1$ and $y_1$ for any $x_1$, and given a $y_1$ there are either two possible solutions, one possible solution, or no possible solution for $t_1$ and $x_1$. But however you decide to determine the values, we want to know all three values $x_1$, $y_1$, and $t_1$.
We also want to know the speed $v_1$ and angle $\theta_1$ of the initial trajectory as it reaches $(x_1,y_1)$. Taking the horizontal and vertical components of the velocity at time $t$ as a function of $t$, we have \begin{align} v_x(t) &= v_0 \cos \theta_0, \\ v_y(t) &= v_0 \sin \theta_0 - g(t-t_0). \end{align} The speed at time $t$ is therefore \begin{align} v(t) &= \sqrt{(v_0 \cos \theta_0)^2 + (v_0 \sin \theta_0 - g(t-t_0))^2}. \end{align} The angle of motion at time $t$ is either $$ \theta(t) = \arctan\frac{v_0 \sin \theta_0 - g(t-t_0)}{v_0 \cos \theta_0} \quad\text{or}\quad \theta(t) = \arctan\frac{v_0 \sin \theta_0 - g(t-t_0)}{v_0 \cos \theta_0} + \pi $$ depending on whether $\cos\theta_0$ is positive or negative, respectively. I'll assume from here on that $\cos\theta_0$ is positive. So at time $t = t_1$, the speed and direction of the projectile are \begin{align} v_1 &= \sqrt{(v_0 \cos \theta_0)^2 + (v_0 \sin \theta_0 - g(t_1-t_0))^2}, \tag1\\ \theta_1 &= \arctan\frac{v_0 \sin \theta_0 - g(t_1-t_0)}{v_0 \cos \theta_0}. \tag2 \end{align}
Now suppose there is a flat surface through the point $(x_1,y_1)$ at angle $\alpha$, where $\alpha$ is the angle between that surface and the positive direction of the $x$ axis; that is, $\alpha = 0$ for a horizontal surface, $\alpha=\frac\pi2$ ($90$ degrees) for a vertical surface, and $\alpha=\frac\pi4$ ($45$ degrees) for a surface at $45$ degrees from horizontal, sloping upward to the right. If we have perfect reflection of a projectile moving in the direction $\theta_1$ as it strikes that surface, the reflected path of the projectile will start at the angle $2\alpha - \theta_1$.
So what we need to do is to reapply the equations for projectile motion, but substitute $2\alpha - \theta_1$ as the new "initial angle", and translate the "initial position and time" from the point $(x_0,y_0)$ and time $t_0$ to the point $(x_1,y_1)$ and time $t_1$. We do this by substituting $x_1$ for $x_0$, $y_1$ for $y_0$, and $t_1$ for $t_0$ in the equation of motion. The result is \begin{align} x &= x_1 + v_1 (t - t_1) \cos (2\alpha - \theta_1), \\ y &= y_1 + v_1 (t - t_1) \sin (2\alpha - \theta_1) - \tfrac12 g (t- t_1)^2, \end{align} where $v_1$ and $\theta_1$ are defined by Equations $(1)$ and $(2)$ above. You can further manipulate these equations algebraically to get them into the exact form you would like.
Alternative method
Another way to derive a formula is to use a little linear algebra. We still need $x_1$, $y_1$, and $t_1$ at the point of impact on the flat surface, but we avoid explicitly computing the speed or angle of the incoming trajectory at that point.
We do still need to think about the velocity of the projectile before and after the collision, but instead of speed and direction, we use the $x$ and $y$ components of velocity, $v_x$ and $v_y$, more directly. It will help to represent the initial velocity in this form as well, that is, set \begin{align} v_{0x} &= v_0 \cos \theta_0, \\ v_{0y} &= v_0 \sin \theta_0 \end{align} so that $v_{0x}$ and $v_{0y}$ be the $x$ and $y$ components of velocity at the start of the first part of the trajectory. The equation of the trajectory can then be written \begin{align} x &= x_0 + v_{0x} (t-t_0), \\ y &= y_0 + v_{0y} (t-t_0) - \tfrac12 g (t-t_0)^2. \end{align}
The components of a velocity vector at a later time $t$ in this part of the trajectory are then $$ \begin{pmatrix} v_x(t) \\ v_y(t) \end{pmatrix} = \begin{pmatrix} v_{0x} \\ v_{0y} - g(t-t_0) \end{pmatrix}. $$ (I have put the components of the vector together in a $2\times1$ matrix here because it is a convenient format for linear algebra. Rest assured that we'll stop using this format when we get to the final answer; I am using it just for the derivation of formulas.)
The next step will be to apply a transformation matrix to reflect this vector around a line parallel to the flat surface, passing through the origin. The reason we want the line to pass through the origin is because vectors have no fixed "starting" position, only a direction and magnitude; but if you imagine the "tail" of the vector at the point $(0,0)$, then the transformation we want will leave the "tail" of the vector in place and only move the "head".
We now have two choices on how to proceed, depending on how we represent the inclination of the reflecting surface.
Using the angle of the surface. If the reflecting surface makes an angle $\alpha$ with the $x$ axis, then the reflection matrix is $$ \begin{pmatrix} \cos(2\alpha) & \sin(2\alpha) \\ \sin(2\alpha) & -\cos(2\alpha) \end{pmatrix} $$ (as shown in this answer to another question), and the reflected vector is found by multiplying the incoming vector by this matrix on the left. The resulting vector is \begin{align} \begin{pmatrix} v_{1x} \\ v_{1y} \end{pmatrix} &= \begin{pmatrix} \cos(2\alpha) & \sin(2\alpha) \\ \sin(2\alpha) & -\cos(2\alpha) \end{pmatrix} \begin{pmatrix} v_{0x} \\ v_{0y} - g(t_1-t_0) \end{pmatrix} \\ &= \begin{pmatrix} v_{0x} \cos(2\alpha) + (v_{0y} - g(t_1-t_0)) \sin(2\alpha) \\ v_{0x} \sin(2\alpha) - (v_{0y} - g(t_1-t_0)) \cos(2\alpha) \end{pmatrix} \end{align}
Writing this as a system of equations without matrices, \begin{align} v_{1x} &= v_{0x} \cos(2\alpha) + (v_{0y} - g(t_1-t_0)) \sin(2\alpha), \tag3\\ v_{1y} &= v_{0x} \sin(2\alpha) - (v_{0y} - g(t_1-t_0)) \cos(2\alpha). \tag4 \end{align}
Using the slope of the surface. If the slope of the reflecting surface is $m$, that is, the surface is parallel to the line $y = mx$, then the reflection matrix is $$ \frac{1}{1 + m^2}\begin{pmatrix} 1-m^2 & 2m \\ 2m & m^2-1 \end{pmatrix}. $$ (See this question and its answer.) The reflected vector is therefore $$ \begin{pmatrix} v_{1x} \\ v_{1y} \end{pmatrix} = \frac{1}{1 + m^2}\begin{pmatrix} 1-m^2 & 2m \\ 2m & m^2-1 \end{pmatrix} \begin{pmatrix} v_{0x} \\ v_{0y} - g(t_1-t_0) \end{pmatrix}. $$ If you do the matrix multiplication, the result is equivalent to the system of equations \begin{align} v_{1x} &= \frac{1}{1 + m^2}((1-m^2)v_{0x} + 2m(v_{0y} - g(t_1-t_0))), \tag5\\ v_{1y} &= \frac{1}{1 + m^2}(2m v_{0x} - (1-m^2)(v_{0y} - g(t_1-t_0))). \tag6 \end{align}
Of course this does not work for a vertical line of reflection, but in that case we simply change the sign of $v_x$ at the point of impact.
Assembling the final result. We already know how to write the equation of a trajectory that starts with given $x$ and $y$ velocity components from a given point at a given time; all we need to do is to take the values of $v_{1x}$ and $v_{1y}$ (either from Equations $(3)$ and $(4)$ or from Equations $(5)$ and $(6)$) and plug them into these equations: \begin{align} x &= x_1 + v_{1x} (t-t_1), \\ y &= y_1 + v_{1y} (t-t_1) - \tfrac12 g (t-t_1)^2. \end{align}
While this may seem like a long procedure, most of the writing above is just explanation and proof; to actually use the procedure, after determining $x_1$, $y_1$, and $t_1$ you merely take four of the equations above (selecting the ones that suit the way your surface is described) and start evaluating them, using the known data.
Move the launch coordinates to $(0,0)$ by setting $x_f=x_f-x_0$, $y_f=y_f-y_0$ and $x_0=y_0=0$.
We can then write the following equations for the projectile's $x$ and $y$ components:
$$y= (v_0\sin \theta) t- \frac{1}{2}gt^2$$ $$x = (v_0\cos \theta) t$$
Let us see at what time $t_1$ the projectile is at the target $x$-coordinate, i.e when is $x=x_f$: $$ t_1 = \frac{x_f}{v_0 \cos \theta}$$ Inserting $t_1$ into the equation for the $y$ component of the projectile, must give the height $y_f$ of the projectile at this time: $$y_f= (v_0\sin \theta) \frac{x_f}{v_0 \cos \theta}- \frac{1}{2}g(\frac{x_f}{v_0 \cos \theta})^2 \tag{1}$$ Remembering that $$\frac{1}{\cos^2 \theta}=1+\tan^2 \theta$$ we can rearrange equation $1$ to become: $$gx_f^2\tan^2 \theta - 2v_0^2x_f \tan \theta + 2v_0^2y_f + gx_f^2 = 0$$ Solving this quadratic equation for $\tan \theta$ gives the formula in the Wiki link I gave in my comment above, i.e. $$\theta = \arctan \left( \frac{v_0^2 \pm \sqrt{v_0^4 - g(gx_f^2+2y_fv_0^2)}}{gx_f} \right)$$
Best Answer
With horizental velocity $v_x=v_0\cos\theta$ and vertical velocity $v_y=-v_0\sin\theta-gt$ (here $\theta>0$) and integration we see $$x=\int_0^tv_0\cos\theta\,dt=v_0\cos\theta\,t\,\,\,,\,\,\,y=\int_0^t v_y\,dy=-v_0\sin\theta\,t-\frac12gt^2$$ Eliminating $t$ between them gives $$y=-\tan\theta\,x-\frac{g}{2v_0^2\cos^2\theta}x^2$$ the range $R$ is the point $(R,y_0<0)$ and you may continue from here!
Edit: The solution for $h=-y_0$ and $\theta<0$ after solving the equation $$\frac{g}{2v_0^2\cos^2\theta}d^2+\tan\theta\,d-h=0$$ is $$d=-\dfrac{v_0^2}{2g}\sin2\theta\left(1+\sqrt{1+\dfrac{2gh}{v_0^2\sin^2\theta}}\right)$$ which works for $\theta<0$ as you wanted.