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What are these variables?

  1. Jan 21, 2007 #1
    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?
     
    Last edited: Jan 21, 2007
  2. jcsd
  3. Jan 21, 2007 #2

    cristo

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    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!
     
  4. Jan 21, 2007 #3
    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.
     
  5. Jan 21, 2007 #4

    cristo

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

    No, I don't know of any other formula.
     
  6. Jan 21, 2007 #5

    D H

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    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.
     
  7. Jan 21, 2007 #6
    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 :(.
     
  8. Jan 21, 2007 #7

    jtbell

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    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.
     
  9. Jan 21, 2007 #8
    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.
     
  10. Jan 21, 2007 #9

    cristo

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    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!
     
  11. Jan 21, 2007 #10
    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.
     
  12. Jan 21, 2007 #11

    cristo

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    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.
     
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