# Mechanics: projectile motion with air resistance

1. Sep 20, 2005

### Dev Null

Hello All,

I was recently intrigued by the site: http://www.lcse.umn.edu/specs/labs/catapult/ which sets about showing a computational solution to the problem of projectile motion with air resistance. I think the site is a great idea, and presented well, but it seems like some of the physics just has to be wrong (their equations are right down the bottom of the page.) Its entirely probable that I'm just misunderstanding it myself, and I'm hoping someone on here can help sort me out. My problem is this:

They solve the problem by dividing the velocity and acceleration into two vectors, and solving for each separately; so velocity v becomes v(x) and v(y), acceleration a becomes a(x) and a(y). Their equation for a(y) reads:

a(y) = (mg - bv(y)^2)/m

where m=mass, g=gravity, and b=drag coefficient. Assuming the usual orientation (positive y is up; negative down) g is presumably a negative number in whatever units pleases you most. But this being the case, then wether the projectile is going up or down (v(y) positive or negative) -bv(y)^2 will always be negative. Which means that air resistance slows the projectile down on its way up - I'm with you so far - and speeds it up on the way back down. Which can't be right, can it?

Can anyone point out where I've got it wrong, or suggest how to fix the equation if I've got it right? Is it just a matter of adjusting the sign of the drag portion of the equation to always oppose the direction of travel?

2. Sep 20, 2005

### krab

You are absolutely right. Their equations are incorrect. They should be:
$$m\vec{a}=m\vec{g}-bv^2\hat{v}$$
($\hat{v}$ is the unit vector in the direction of v.) So the y-component is:
$$ma_y=mg-b|v|v_y$$

I suggest you inform the people running the website. A clear nonphysical result of their incorrect equations is that the trajectories with and without air resistance cross each other.

3. Sep 21, 2005

### Dev Null

Nifty Latex equation thing, let me see if I can manage it; should the following:

$$ma_y=mg-b|v_y|v_y$$?

Or is it correct that the unvectored velocity (if you'll pardon me for making up a term) should get worked back into the y-component?

Its not that I'm trying to nitpick, but my understanding of the area is fragile enough that while I expect it should be the y-component only, I'm really none too certain... Thanks again,

Last edited: Sep 21, 2005
4. Sep 21, 2005

### krab

No. Look again at the vector equation. Realize that $\hat{v}=(v_x/|v|)\hat{i}+(v_y/|v|)\hat{j}$.

5. Sep 21, 2005

### Dev Null

Aha. Got it now. Thanks.

You've been so helpful I might be so bold as to ask the question that got me started on the projectile/air resistance problem in the first place, because I'm not having much luck on my own. I have a water rocket whose behaviour I'm trying to explain with no more instrumentation than a pocket watch, a protractor, and a measuring tape. What I'd like to be able to do is work backwards from the performance of the rocket and determine the launch velocity and the coefficient of friction for the rocket. I had thought that I should somehow be able to work this out by shooting straight up and measuring the relationship between the time from launch to apogee, and the time from apogee to landing. If I've worked it out correctly the first should be shorter than the second so long as the launch speed is faster than the terminal velocity of the rocket (and I think I can count on that) and I thought perhaps the difference would be indicative of some relationship between the two unknowns (velocity and friction). I can get additional information by firing shots at various angles and measuring the distance they travel or their flight time, though practically timing the apogee is much harder in this case.

Any ideas? Do I have enough information to find both unknowns? Or am I stuck trying to find some way to measure the initial velocity?

6. Sep 21, 2005

### SGT

You should measure not only the times to apogee and landing, but the x and y coordinates for each one. Since you have two unknowns: initial velocity and drag coefficient, those measurements should be enough to estimate the values of the unknowns.
Of course those are very poor informations. In order to estimate the values with any accuracy, you would need several times and coordinates along the trajectory.
Also, remember that your measurements have errors, so you should repeat them several times.

7. Sep 21, 2005

### krab

If you have access to a sufficiently tall building and a radar gun, you can get the terminal speed independently.

8. Sep 21, 2005

### Dev Null

Well the purchase of the radar gun is part of what I was trying to avoid - plus I rather like doing experiments in mechanics with only the tools that Newton would have had to work with (give or take the fact that my stopwatch is digital.) But I suspect that I could rig up a bit of a sighting scope with the protractor to measure the height at apogee SGT, and maybe from that and the horiziontal distances at various angles I can work something out.

And yes, I'll have to repeat the experiment quite a few times to get anything like accuracy, but the rocket seems to be remarkably consistent so long as I use the same pressure, which I can measure.

9. Sep 22, 2005

### SGT

If you could have a closed form equation, there would be straightforward to determine the two unknowns by replacing in the equation the x and y coordinates at two distinct time instants.
The nonlinearity of the problem makes impossible to find a closed form equation, so you must work iteratively: Make educated guesses for the values of $$v_0$$ and b and find the resulting trajectory, working with small time increments (Matlab is just fine for that). Compare your theoretical results with your measurements and use the error to correct your guesses. Repeat until you have a good approximation for your values. Only them shall you make another experiment in order to reduce the measurement errors.