# Projectile motion & drag

1. Feb 19, 2007

### Hypochondriac

Ok so (UK) A2 Physics coursework time came, and I stupidly thought that projectile motion would be a fun investigation. But after just a little research I found it most challenging, it's taken me 'round the houses' via Reynolds numbers etc, and I've come to a halt as I'm kinda stuck.

I plan to see any relationship between the ideal, negligable-air-resistance trajectory of the projectile and its actual flight path. Perhaps compiling my own equation for the real range of the projectile, as an adaptation of the negilable air resistance equation:

$$x=\frac{2v^2\cos\theta\sin\theta}{g}$$
therefore
$$x=\frac{v^2\sin2\theta}{g}$$

My first question is more of a technical one that a calculation queery; would my capture method work?
I plan to set up a camera with a long shutter exposure in a very dark room. I would then set off a stobe light that would have a definate known flash frequency and I'd fire my projectile at the force and angle I wish. As the projectile (mainly spherical objects, perhaps other shaps when doing further investigation) takes it's flight, the light from the stobe reflects off of it and exposes it to the camera. Hopefully the light wont reflect much off of the dark background and will show the ball in it's flight at set times, enabling me to calculate initial velocity and the such.
So I'll be greatful for any comments on this.

Next, upon research I found that in most cases air resistance is assumed negligable which I feel would not give me much to write about and give my coursework an "AS" feel to it. So I am including it to give the investigation more depth.

I found that air resistance is sometimes proportional to velocity but it is more realistic to look at it as a square law, which also helps in gaining A2 marks.
As I learnt from AS, projectile motion can be split into two componants, verticle and horizontal.
First off, the horizontal velocity would've been constant, but with an external force on it (drag) it will -negatively- accelerate and therefore won't be constant.
But as the projectile takes it's flight path diagonally up, across, then diagonally down, so the drag is always opposing is diagonally down then across, then diagonally up. (see fig 1. attached)
As the angle is constantly changes obviously the cosine of it constantly changes, therefore the horizontal componant of the drag (R in my diagram) changes, therefore it's non uniform acceleration...which as far as I'm aware may require some differential equations, I don't know how these are set up or used so I may need some help with that if possible.

As for the verticle componant of my projectiles trajectory, I split that up into two. Up and down,
For up, (as far as I'm aware)
initial velocity = $$v\sin\theta$$
final velocity will be 0 at y(max)
acceleration either encounters the same problems as the verticle acceleration, or perhaps behaves as follows
$$\left(\frac{\frac{1}{2}\rho v^{2} A C_{d}}{m}\right)-g$$
where rho is the density of air, v is the velocity (which of course is constantly changing due to this force so i don't know how that will work), A is the reference area, which apparently is related but not equal to the cross sectional area (not sure how to find that) and Cd is the drag coefficient.
Although I'm uncertain on how correct that formula is or whether it's applied in this situation.

As for the return flight down to earth:
initial velocity would be 0
final velocity would be the objects terminal velocity which is given by the formula
$$\sqrt{\frac{2mg}{\rho A C_{d}}}$$
correct? or would it not be it's terminal velocity?
and the acceleration would be the reverse of before
$$g-\left(\frac{\frac{1}{2}\rho v^{2} A C_{d}}{m}\right)$$

Overall, the time taken will be the time taken for both parts of the verticle motion, and the range will be related to the time taken and the velocities/deceleration of the horizontal motion.

I'm unsure whether these equations are accurate or precise, and my main worry is how this differential equation thing is going to turn out. It's the only way I see around this constantly changing drag direction.
I'm not even entirely sure what im showing/finding/proving if I'm honest

So please tell me any comments, or anything ive done wrong or anything really
(and if you're wondering most of my background info came from wikipedia)

hope I ain't violated any rules, sorry about any spelling or latex errors, look forward to hearing from you lot, thanks in advance. ^_^

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Last edited: Feb 19, 2007
2. Feb 20, 2007

### AlephZero

Yes your capture method should work. Using high speed photography is a standard way to measure what is happening in this type of situation. In my work I've used speeds up to 10,000 pictures per second, but you probably don't need to go that fast!

You need to take some photos so you can convert the size of the image on the photo into distances in meters, of course. Remember the distance from the camera lens to the object may vary depending whether it is in the centre of the picture or at the edge. One way to allow for that would be to make a large piece of "graph paper", put it in the plane where the object will move, and photograph it.

For different objects of the same size and shape moving at the same speed, the force created air resistance will be (approximately!) the same. You can change the effect of the force on the motion by changing the mass of the object. Imagine trying to kick a football-sized balloon, compared with kicking a real football, for example.

You can simplify the equations so you can measure the actual drag force on your object at different speeds, by making the horizontal velocity zero. Then you could those results to predict the general motion.

Hope this helps - good luck.

3. Feb 20, 2007

### Hypochondriac

In my capture method I wasn't going to take multiple concecutive shots though, i hoped the long exposure would give me one picture of the ball in many positions.

I kinda understand what you mean about making the horizontal velocity 0, but wouldnt I need to know the affect drag has on the horizontal velocity as I'm mainly working with range.
And I'm pretty sure that somewhere I read said I need to have some sort of differential equation, seen as the drag direction is changing therefore the drag in the horizontal is changing (w.r.t. the y=cosx graph)

4. Feb 20, 2007

### AlephZero

Yes, that should work OK. The flashes from a strobe are very short duration so each image should be sharp.

Of course it's much easier to work with the images and measure positions etc, if you use a digital camera, compared with old-style photography.

Assuming you are using a spherical object, and it's not spinning, the drag force should depend only on the speed, and act in the opposite direction to the velocity, whatever that direction is. If you start off by dropping the object vertically you know there are only two forces on it (weight and drag) and you can measure its velocity and accleration and so find how the drag force varies with speed. You should be able to compare the experiment with theoretical predictions as well (Look for data on drag coefficients as a function of Reynolds number).

Yes you will need to set up a differential equations for the "range" motion, but you might not be able to solve them analytically. If that's the case, you could solve them numerically with a computer - start at time 0, work out the horizontal and vertical forces, assume the forces stay constant for a short time (maybe 0.01 sec) and work out the position of the object using the "x = vt + at^2/2" equations, then repeat for the next 0.01 sec. You could program this in a spreadsheet, or in a language like C or Basic.

When you do experiments like this, you need to check out that things work properly one step at a time, by building up from something simple. Starting with a vertical drop of a heavy object and verifying that g = 9.8 m/sec would be a good thing to try first. Then add in some air resistance, then go for the horizontal and vertical motions combined. As you do the simple experiments, you will learn the best way to do the more complicated ones.