How to determine where an object will land on a variable elevation surface? I'm making a game and I need to figure out the exact point on a hilly surface where the car will land after the jump.
Here is an image:

So since this is a game, it's very easy for me to figure out the start position, velocity angle and speed of the car. However, I'm not sure how to calculate many points along the curved line the car will follow as it falls to the ground. All the examples I've been able to find either assume the ground is flat, or assume I know what the final elevation will be.
Basically, what I have been unable to find is a formula where I can put in a X distance and figure out what the Y is on that curve. If anyone could point me in the right direction, that would be incredibly helpful.  
 A: You're given $v$, $\theta$, $x_0$, and $x_y$ and you want to calculate $y$ in terms of $x$. Then you can use that to plug into the equation for the surface and solve.
Firstly, resolve $v$ into $v_x = v\cos(\theta)$ and $v_y = v\sin(\theta)$.
Then $x = v_xt$ and $y = v_yt+\frac{1}{2}gt^2$ (assuming you have a constant gravitational field which isn't really a given I suppose!)
Then eliminate $t$ to give:
\begin{equation}y = \frac{v_y}{v_x}x + \frac{gx^2}{2v_x^2}
\end{equation}
Of course, $\frac{v_y}{v_x} = tan(\theta)$, so you can put that in if you like. If you have a different $g$ you'd have to do the integration to figure out the equation for $y$. And if $g$ had an $x$ component you'd need to integrate it into the equation for $x$. 
Anyway, you'll notice that we have a parabola. And since $g$ is negative, it's a frowny parabola which is what we would expect.
Now it depends on what your elevation is. If you have a nice equation for that surface you can solve them simultaneously. Depending on details, you may have more than one solution (for example, if your path goes through a hill, there might be two solutions: one for each side). Normally, the first solution would be what you're looking for, so the one with the highest $x$ (or lowest depending on your axes and direction of movement, but assuming normal conventions and the picture you showed, it would be highest).
