Idealized trajectory from sloped surface

Question

I am a GIS programmer implementing a visualization.

I am modeling the idealized trajectory of a particle ejected from a volcanic vent using:

$$\text{distance} = \frac{(v^2 \times sin(2\theta))}{g}.$$

Where $g = 1.62\:\mathrm{m/s^2}$, $v$ is velocity, and $\theta$ is ejection angle. $g$ is the lunar gravity constant I was supplied.

How can I incorporate the slope of the underlying surface assuming a single point of ejection? $$$$$$$$

EDIT: My current workflow is to compute total travel distance, extract a topographic profile along the total theoretical travel distance and then check the height of the projectile to the height of the actual surface at 100m intervals. In this way I can compute the landing site for the projectile.

EDIT 2: I updated the question with the correct formula. Apologies for the incorrect transposition. My implementation now assumes a completely flat surface. What happens when the ejection surface is sloped either uphill or downhill?

Answer 1

An idealized projectile launched from ground with speed $v$ at angle $\theta$ measured from zenith will reach ground at time:

$$ t=\frac{2v\sin\theta}{g} $$

assuming the ground is level. The horizontal distance it will have traveled in that time is:

$$ d=\frac{2v^2\sin\theta\cos\theta}{g}=\frac{v^2\sin 2\theta}{g} $$

For small $\theta$ you could approximate this with:

$$ d\approx\frac{2v^2\theta}{g} $$

but I guess you need a formula that works also for larger $\theta$.

If the ground is sloped with a constant angle $\alpha$ (positive angles for uphill slopes) from the ejection point to where it hits the ground, you can use the following formula:

$$ d=\frac{2v^2\sin\theta(\cos\theta-\sin\theta\tan\alpha)}{g} $$

Do you want me to show you how I derived the formula?