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A simple problem involving motion in two directions

  1. Oct 15, 2007 #1
    This is my first post, so an introduction is in order.

    My educational background includes a MBA from UGA in 2007, a MS in EE from Southern Illinois U in 1994, and a BS in EE with minors in physics and math from SIU in 1992.

    Recently I've applied to three universities for their PhD physics program. And I have to take the GRE Physics subject exam.

    I've been studying using an old college physics text book by Ohanian. It's an old book, probably 20 years old or so. But I've been going through each chapter working the even numbered problems (because the answers are in the back). One particular problem has thrown me for a loop. Here it is...

    Here are some equations:

    Vx = V cos(θ)
    Vy = V sin(θ) - gt

    x=Vt cos(θ)
    y=Vt sin(θ) - ½gt²

    I've been trying to solve this problem by writing y in terms of x and v, but I can't get rid of θ.
  2. jcsd
  3. Oct 15, 2007 #2


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    Just work with the last two equations (for x and y). You have two unknowns (t an dtheta) for two equations so it's solvable. To getrid of the angle, isolate sin and cos and then use that sin^2 + cos^2 = 1. That will leave you one equation containing x, y and t. You can therefore solve for t.
  4. Oct 15, 2007 #3
    Well, I get a mess that looks like this:

  5. Oct 15, 2007 #4


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    you mean 1/4 for the first term.

    Now plug in th evalues of x and y and g and solve for t.
  6. Oct 15, 2007 #5
    evalues? It's been almost 20 years since I've done stuff like this. Please clarify.
  7. Oct 15, 2007 #6


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    I think that it's clear from my message that it's a typo, it should be "the values" instead of "th evalues"
    You have an equation that contains t^4, t^2 and a cinstant. You can solve for the value of t^2 using the quadratic formula (in other words, just rename t^2=w, say. Then you have an equation of the form a+bw + cw^2 =0 with a,b,c being some constants. Use the quadratic formula to find w, which is t^2. )
  8. Oct 15, 2007 #7
    I see. I thought you might have been talking about eigenvalues.
  9. Oct 22, 2007 #8
    Need help solving this for θ:


    I've tried just about every trig identity known to man.

  10. Oct 23, 2007 #9
    a sec²θ + b tan θ - c = 0

    use Pythagorean identity sec²θ = tan²θ + 1

    a(tan²θ + 1) + b tan θ - c = 0

    a tan²θ + b tan θ + (a - c) = 0

    Use of the quadratic formula yields

    tan θ = [-b ± √(b² - 4a(a - c))]/2a

    θ = arctan{[-b ± √(b² - 4a(a - c))]/2a}
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