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Need some help solving for 'v' in this equation

  1. Jul 13, 2011 #1
    given this formula for the range of a projectile when the initial height (y) is not zero:

    Code (Text):
    d = (v * cos(a) / g) * (v * sin(a) + sqrt((v * sin(a))[SUP]2[/SUP] + 2 * g * y)
    in a project i'm working on i need to compute 'v' given 'd', 'a', 'y' and 'g', so i would like to rewrite this equation in terms of 'v' - my trig identities aren't so hot, could someone possibly help me to express this equation in terms of 'v' instead of 'd' ?

    TIA

    *er - this was supposed to post in General Math...oops. i dunno how to move it there w/o crossposting..
     
    Last edited: Jul 13, 2011
  2. jcsd
  3. Jul 13, 2011 #2

    micromass

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    Hi xovangam! :smile:


    So the equation is

    [tex]d=\frac{v\cos(a)}{g}(v\sin(a)+\sqrt{v^2\sin^2(a)+2gy})[/tex]

    Right?

    The square root is the most important problem here. So you'll first going to need to rewrite your equation as

    [tex]\frac{gd}{v\cos(a)}-v\sin(a)=\sqrt{v^2\sin^2(a)+2gy}[/tex]

    and square both sides. You'll end up with a quadratic equation in v.
     
  4. Jul 13, 2011 #3
    is this on the right track ?

    Code (Text):
    ((gd/v*cos(a)) - v*sin(a))^2 = v^2*sin(a)^2 + 2gy

    A = gd/v*cos(a)
    B = v*sin(a)

    (A - B)(A - B)
    = A^2 - AB - AB + B^2
    = A^2 - 2AB + B^2

    (gd/v*cos(a))^2 - 2*(gd/v*cos(a))*(v*sin(a)) + (v*sin(a))^2 = v^2*sin(a)^2 + 2*g*y

    => (gd)^2/(v*cos(a))^2 - (2*gd*v*sin(a))/(v*cos(a)) + v^2*sin(a))^2 = v^2*sin(a)^2 + 2*g*y

    => (gd)^2/(v^2*cos(a)^2) - (2*gd*sin(a))/(cos(a)) = 2*g*y
    how do i get v^2*cos(a)^2 out of the bottom of the fraction on the left, multiply both sides by v^2*cos(a)^2 ? it seems like i'm going to end up with 'B' = 0 in the simplified equation ? (i.e. Av^2 + Bv + C)
     
  5. Jul 13, 2011 #4

    micromass

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    Looks good. So now have something of the form

    [tex]\frac{A}{Bv^2}+C=D[/tex]

    This is very easy to solve for v! Indeed:

    [tex]\frac{A}{Bv^2}=D-C[/tex]

    Thus

    [tex]\frac{Bv^2}{A}=\frac{1}{D-C}[/tex]

    Hence

    [tex]v^2=\frac{A}{B(D-C)}[/tex]
     
  6. Jul 13, 2011 #5
    cool. so

    Code (Text):
    A = (g*d)^2
    B = cos(a)^2
    C = -(2*g*d*sin(a))/(cos(a))
    D = 2*g*y
    and thus

    Code (Text):
    v^2 = (g*d)^2/(cos(a)^2*(2*g*y + (2*g*d*sin(a))/cos(a))
    ?
     
  7. Jul 13, 2011 #6

    micromass

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    Seems right!
     
  8. Jul 13, 2011 #7
    and can i further simply that to:

    Code (Text):
    v = (g*d)/cos(a) * sqrt(2 * g * d * tan(a) + 2 * g * y)
    ?
     
  9. Jul 13, 2011 #8

    micromass

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    If you know that everything is positive, yes.
     
  10. Jul 13, 2011 #9

    Ray Vickson

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    No. You will end up with a quadratic equation in v^2, so if you set y = v^2 you get a quadratic in y.

    RGV
     
  11. Jul 13, 2011 #10

    micromass

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    If you check the solution in the previous post, you see that you do end up with a quadratic equation in v. Is the solution wrong somewhere?
     
  12. Jul 13, 2011 #11

    Ray Vickson

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    The equation you wrote before had the form
    A/v - Bv = sort(v^2 + c), so squaring it gives
    A^2 /v^2 - 2AB + B^2 v^2 = c + v^2. Multiply through by v^2 to get a quadratic in v^2.

    RGV
     
  13. Jul 13, 2011 #12

    micromass

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    And in this case, it was a quadratic. Like I claimed. I don't see your point.
     
  14. Jul 13, 2011 #13

    Ray Vickson

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    I was responding to the statement that it is quadratic in *v* (not v^2). Of course, that might have been a typo.

    RGV
     
  15. Jul 13, 2011 #14

    micromass

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    Really, excuse me, I don't understand what the confusion is here. If you square both sides of the equation and rework it a bit, then the resulting equation will be quadratic in v, no??
    This is certainly not true with every such equation (where you indeed have to make a substitution), but it is true with this equation, and that's all I said...
     
  16. Jul 13, 2011 #15

    Ray Vickson

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    The equation becomes A^2 - 2AB v^2 + B^2 v^4 = c v^2 + v^4, where B is not equal to +1 or -1, so the v^4 terms do not cancel.

    RGV
     
  17. Jul 13, 2011 #16

    micromass

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    In this case, they do! So in this case, we do have a quadratic equation. (or did I make a mistake anywhere??). I really don't see what the big deal is here.
     
  18. Jul 13, 2011 #17

    Ray Vickson

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    You are right! Sorry: I was looking at the wrong equation.

    RGV
     
  19. Jul 14, 2011 #18
    In brief :
     

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