EDIT: jesus I gotta learn to read properly, lol. Under construction.

EDIT: OK, constructed. My apologies for not reading your post properly in the first place.

So the bullet has a speed of 5 units per second. I recommend changing this into a SI unit measurement like meters per second for example.

EDIT: You want to replace that simple increase with a check to see how many seconds have passed since your last frame (we will call this t), and increase your bullet's position by t*speedX. EDIT: the reason for that is that otherwise your bullet's speed will depend on your framerate.

Now, onto your actual question. Air resistance is complicated at high speeds. For now lets look at a simplification of it, meaning low speeds and low viscosity mediums like air is. (EDIT: simple physics is usually more than fine for games). The resistive force on the bullet is:

F = -bv

where v is the [s]speed[/s] velocity of the bullet, and b is the drag coefficient.

we know v, so we need to know b to get F. Assuming our bullet is round:

b = 6 * Pi * n * r

where the symbol like n is the "dynamic viscosity" of the air and r is the radius in meters of the bullet. You must decide on the radius of the bullet, but n is 18.6 * 10^-6 Pa s for air at room temperature.

This tells you the force on your bullet.

F = ma, so by setting the mass of the bullet you can calculate the deceleration on the bullet due to air resistance. This will tell you by how much to update speedX for the next frame.

EDIT: but you will want to update F every frame too to account for the changing speed.

I didn't read your original post.

No matter. I edited it because I hadn't read your question correctly. I now hope my above, edited post helps to answer your OP.

Thanks for the help. I really appreciate it.

But its very complicated.

You're welcome.

OK, here's a simpler version of my solution:

The simplified acceleration of the bullet due to air resistance is:

a = c * r * speedX / m

where r is the radius of the bullet, m is the mass of the bullet, and c is a constant you can play with till the movement of the bullet looks right.

EDIT: in fact if you don't really care about the mass and radius of the bullet you can just use:

EDIT: a = k * speedX

EDIT: where k is a constant you can play with till the movement of the bullet looks right. That's actually a much easier way to do it because you won't have to worry about units at all. The downside is you can't take into account different sized or different weight bullets very well, and you can't take into account different movement mediums like air or water...

So every frame you need to do the following:

1 - find out how much time t has passed since the last frame (zero if its the first frame)
2 - change the position of the bullet by t * speedX
3 - calculate a
4 - change speedX by a * t

1 - find out how much time t has passed since the last frame (zero if its the first frame)
2 - calculate a
3 - change speedX by a * t
4 - change the position of the bullet by t * speedX

Assuming you're doing variable framerate physics, which is generally a bad idea.

[quote name='forsandifs' timestamp='1305119518' post='4809399']
1 - find out how much time t has passed since the last frame (zero if its the first frame)
2 - calculate a
3 - change speedX by a * t
4 - change the position of the bullet by t * speedX

Assuming you're doing variable framerate physics, which is generally a bad idea.
[/quote]

Hmm, the solution I proposed is independent of framerate. Why do you think physics that is independent of framerate is a bad idea?

EDIT: I think physics that is independent of framerate is the only way to go. Physical equations depend on units, and if your units of time are unpredictably variable as is the time taken for every frame, then your physical laws will also be variable and unpredictable, which is not a good thing.

EDIT: there is a weakness in the solution however, in that it becomes less accurate the lower the framerate is. But to solve that would require formulating x with respect to t, which I can't quite figure out right now.