What Do the Variables in the Projectile Trajectory Equation Represent?

[randomvictim]

http://upload.wikimedia.org/math/4/2/f/42fa657994ca819eccfcf2b36296ddf9.png

Sorry I can't display images on these forums.

That is the equation of finding the trajectory of a projectile with gravity and air resistance.

I know that

m = mass
vº = initial velocity
e = 2.71828
g = -9.81m/s²

What do those other variables stand for?

 

[cristo]

Well, clearly t is time. I'd say that k was probably some constant of proportionality to do with the air resistance, but since I don't know where you got this equation, I cannot say for sure!

 

[randomvictim]

I got it from the bottom of

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile

If you know of another formula for finding the trajectory of a projectile with gravity and air resistance then that would be great.

 

[cristo]

In the paragraph under the diagram it says F_air=-kv, so k is a constant of proportionality.

No, I don't know of any other formula.

 

[D H]

Note well: That entire section of that article is incorrect. See the discussion page. As is noted on the wiki article's discussion page, air drag is proportional to the square of the velocity. The derivation starts with the wrong equations of motion.

 

[randomvictim]

did it. Look how ridiculous this formula is.

m = mass
v = initial velocity
ø = angle
e = 2.71828182846
k = air constant
g = -9.81 m/s²(((m)(v)(sin(ø))/k)(1-e^(-((k)(t))/m)+(((m^2)(g))/(k^2))(1+(((k)(t))/m)-e^(-((k)(t))/m))

edit: it's wrong?!? I refuse to believe this is wrong after I spent all that time working on it :(.

 

[jtbell]

For low speeds, air drag is roughly proportional to the speed. For higher speeds, it's roughly proportional to the square of the speed. The "critical speed" where drag shifts from one formula to the other depends on the size of the object.

 

[randomvictim]

Well I am not really shooting potatoes, they are frozen grapes, cut down to have a diameter of a half inch. The front of the grape is fairly rounded and sort of reduces on air resistance but I am looking at speeds well over 150 feet per second coming out of the barrel, higher speed in my opinion but low and high are vague terms.

 

[cristo]

Note well: That entire section of that article is incorrect. See the discussion page. As is noted on the wiki article's discussion page, air drag is proportional to the square of the velocity. The derivation starts with the wrong equations of motion.

For a projectile, the simplest way to include air resistance is to consider it as being proportional to the velocity. If we assume the projectile travels at a low speed, then this is a reasonable model.

edit: Didn't see jtbell's post!

 

[randomvictim]

cristo, I know what you're saying and I knew how to calculate drag that way, but calculating the drag at a given velocity doesn't do me any good, it needs to be in a y= format for me to be able to graph it. I couldn't figure out how to derive that from just drag force.

 

[cristo]

cristo, I know what you're saying and I knew how to calculate drag that way, but calculating the drag at a given velocity doesn't do me any good, it needs to be in a y= format for me to be able to graph it. I couldn't figure out how to derive that from just drag force.

I wasn't saying you should calculate drag at a given velocity. I was responding to D H, and saying that the orginal formula from which your expression for distance is derived, is valid in low speed situations.