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Gravity as a Function of X

  1. Dec 14, 2007 #1
    This isn't a homework question, just one that's been nagging me for a while, and it is algebra so I thought this an appropriate place for it.

    I've been working on this problem at my whiteboard for a good half hour and can't figure out. I'm sure it's very simple but a search of Google and these forums uncovered nothing of any value.

    I want to find the point at which an object's velocity will reach 0 if its initial velocity is known and gravity acts at its normal [tex]-9.8m/s/s[/tex]. The problem is that I don't know how to express this as [tex]f(x)[/tex] or [tex]y[/tex]. I've read that gravity is [tex]9.8m/s^2[/tex] but I really don't understand how that works.

    I apologize for asking such a basic question, I just don't remember anything from the last quarter of my algebra class. Can anyone help me out? Thanks in advance.

    I went a little overboard with the tex thing. Oh well. [tex]:)[/tex]
     
  2. jcsd
  3. Dec 15, 2007 #2
    how is this an algebra question? there are very simple formulas out there for this problem.
     
  4. Dec 15, 2007 #3

    HallsofIvy

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    You are saying that the body's acceleration (derivative of velocity) is -g.
    [tex]\frac{dv}{dt}= -g[/tex] so, integrating, v= -gt+ C. Taking the "initial speed" to be v0, v(t)= -gt+ v0. To determine when the objects velocity is 0, set that equal to 0 and solve for t: -gt+ v0= 0 so -gt= -v0 and t= v0/g.

    Since this is neither abstract nor linear algebra, I am moving it.
     
    Last edited: Dec 15, 2007
  5. Dec 15, 2007 #4

    Gib Z

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    My answer: when x=0, assuming the object is inelastic.

    EDIT: Ahh nvm you meant on the objects way up, I thought you meant after it has already started falling.
     
  6. Dec 15, 2007 #5
    Thanks for the help.

    I thought this was an algebra question, sorry it was in the wrong place. Shows how little I know about math.
     
  7. Dec 15, 2007 #6
    It essentially algebra, just not linear algebra/abstract algebra
     
  8. Dec 15, 2007 #7
    The Answer

    ok, I'm assuming you mean that you are shooting an object straight up with some initial velocity. I'd use conservation of energy to solve this.

    at the point where v=0 all of our initial kinetic energy is converted into gravitational potential energy:
    [tex]\frac{1}{2}mv^{2}=mgy[/tex]

    we solve for y:
    [tex]y=\frac{v^{2}}{2g}[/tex]

    And there you have "the point at which an object's velocity will reach 0". It is important to note that this only works when gravity really is 9.8 m/s^2. If you're shooting things off into space then you must take into account that g is not constant so it gets a bit more complicated.
     
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