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How to solve for v?

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    theta = tan^-1 {{v^2 - [v^4-g(gx^2)]}^1/2}/gx
    I know that x is 30, theta is 45, g is 9.81 m/sec^2.
    I'm trying to solve for v.

    2. Relevant equations

    3. The attempt at a solution
    I've gotten all the way down to
    v^2 = (v^4-86612.49)^1/2 + 4.165
    What do I do from here?
    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 13, 2010 #2


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    Homework Helper

    you'll need to re-write it as

    v2 - 4.165= (v4-86612.49)1/2

    then square both sides and solve. You may need to use a numerical method. You'll need to test your solutions as squaring both sides might give you false answers.
  4. Feb 14, 2010 #3
    Once you square it you will be left with only 1 V term. It may be easier to replace your constants with variables. Then just substitute them in once you find V.
  5. Feb 14, 2010 #4


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    Science Advisor

    How "only 1 V term"? There will be [itex]v^4[/itex] and [itex]v^2[/itex] terms but then you can let [itex]u= v^2[/itex] and will have a quadratic to solve for u.
  6. Feb 14, 2010 #5
    By squaring both sides, I get v^2 - 17.35 = v^4 - 86612.49.
    That gets me -17.35 = -86612.49.
    Is this what you mean by a false answer?
  7. Feb 14, 2010 #6
    If i do that, I end up with u = v2 + 416.2
    then the discriminant will end up negative leaving me with an imaginary velocity. Did I do this correctly?
  8. Feb 14, 2010 #7
    If i do that, I end up with x = 86612.49 and y = 4.165. I then end up with v^4 - y^2 = v^4 - x, which again gets me to -y^2=x.
  9. Feb 14, 2010 #8
    v^2 -4.165 = (v^4-86612.49)^1/2
    [tex] v^2 - x = (v^4-y)^{1/2} [/tex]
    [tex] v^4 - 2v^2x + x^2=v^4-y[/tex]
    [tex] -2v^2x + x^2=-y[/tex]
    [tex] v=\sqrt{\frac{x^2+y}{2x}}[/tex]
    there u go
  10. Feb 14, 2010 #9
    The left-hand side is not squared properly. The left hand side should be [itex](v^2 - 17.35)^2 = v^4 - 2(17.35)v^2 + (17.35)^2[/itex]. The v4's will then cancel and you will be left with a quadratic in v. You will need to check the two answers you get, as squaring both sides of an equation introduces new solutions that may not solve the original equation. Ie., x = -1 and x2 = 1, which introduces the solution x = 1.
  11. Feb 14, 2010 #10
    Thank you so much!
    I get it now. I totally squared the left wrong.
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