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Projectile Motion Equation Need Help Re-arranging Please!

  1. Mar 2, 2009 #1
    Hey, this is my first time posting here, I just found this site in a Google search.

    I've been doing a projectile motion question, and I know what I am doing and solving for and such, but I am having great difficulties re-arranging the equation I need. The equation looks like so:
    [tex]\Delta[/tex]d[tex]_{}y[/tex] = v[tex]_{2y}[/tex][tex]\Delta[/tex]t + [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]

    The variable I need to solve for is time ([tex]\Delta[/tex]t), but I'm having vast difficulties in doing so. I know I've solved this before, but am not entirely sure how I did it. If this makes no sense, I will post the whole question so I can be told what I'm doing wrong.

    Thanks in advance for the help, I would really appreciate it!

    EDIT: I think I'm doing this posting thing all wrong...sorry for my noob-ness.

    Ninjaedit: the speed shouldn't be squared, it should be v2 (as in second) y...subscript doesn't work for me properly apparently

    EDIT AGAIN: Okay, so in reading around a little bit more, I noticed I should list what I went through to try to rearrange this equation.
    1. [tex]\Delta[/tex]d[tex]_{}y[/tex] - v[tex]_{2y}[/tex][tex]\Delta[/tex]t = [tex]\frac{1}{2}[/tex]g[tex]\Delta[/tex]t[tex]^{2}[/tex]
    2. [tex]\frac{[tex]\Delta[/tex]d[tex]_{}y[/tex] - v[tex]_{2y}[/tex][tex]\Delta[/tex]t}{0.5g}[/tex] = [tex]\Delta[/tex]t[tex]^{2}[/tex]

    And that's where I got stumped. How would I isolate this variable? Is it do-able?
     
    Last edited: Mar 2, 2009
  2. jcsd
  3. Mar 2, 2009 #2

    Nabeshin

    User Avatar
    Science Advisor

    'Ello, and welcome to PF!

    In general, the solution to this is a quadratic which, depending on your math background, can look really sticky or really neat. If you re-write the equation like this:
    [tex]\frac{1}{2}g\Delta t^{2}+v_{2y}\Delta t - \Delta d_{y}=0[/tex]
    You can see it's in the form of a quadratic...
    [tex]at^{2}+bt+c=0[/tex]

    And I'll assume you're familiar with the quadratic formula...

    Edit:

    There is a special case of this when [tex]\Delta d_{y}=0[/tex] which isn't a quadratic (well it is, but one solution is always t=0).
     
  4. Mar 2, 2009 #3
    Aaah, thank you very much for the reply! And yes, I am familiar with the Quadratic Formula, but haven't used it in a year. I made the mistake of stopping math at Grade 11 U level, in the same semester as my Grade 11 U Physics, and then taking Grade 12 U Physics, after having no math for a year.

    tl;dr what's the Quadratic equation again?

    Thanks again!

    EDIT: Nvm, got it, thanks!
     
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