Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
If you want to know the trajectory, what you are interested should be x(t) and y(t) instead of distance. Also what is the meaning of the last questions? n points is no different from a single point. Can you clarify that –  hwlau Jan 16 '13 at 18:56
1  
One more things, your equation looks wrong and g, how do you get this? –  hwlau Jan 16 '13 at 18:58
    
@Jay Laura Firstly, your formula is still wrong. $\frac{v^2\sin2\theta}{g}\neq\frac{2v^2\sin\theta}{g}$ Secondly, I've updated my answer with a formula that might help you. –  jkej Jan 17 '13 at 9:28

1 Answer 1

up vote 1 down vote accepted

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?

share|improve this answer
    
I updated my question to hopefully include more information. Apologies for an incomplete question. Quite out of my element! –  Jay Laura Jan 17 '13 at 2:31
    
Thanks. I would be interested in knowing how the derivation works. Again, not a physicist, but interested. –  Jay Laura Jan 17 '13 at 13:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.