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Mechanics/Projectiles/Angles/Trig. Identities?

  1. Sep 29, 2007 #1
    1. The problem statement, all variables and given/known data
    [​IMG]

    Apologies for the scrappy diagram. My MS Paint skills aren't amazing.


    2. Relevant equations

    Personally, I'm not too sure. This question (I think) involves things that I haven't studied, I tried to do a little research into it on the Internet but it didn't help too much in the end. All I could really think of was

    sec [tex]\theta[/tex] = 1/cos[tex]\theta[/tex]

    tan^2 [tex]\theta[/tex] + 1 = sec^2 [tex]\theta[/tex]

    sin^2 [tex]\theta[/tex] + cos^2 [tex]\theta[/tex] = 1

    tan^2 [tex]\theta[/tex] = (1 - cos2[tex]\theta[/tex])/(1 + cos2[tex]\theta[/tex])

    cos2[tex]\theta[/tex] = (2cos^2[tex]\theta[/tex] -1)

    And maybe Pythagoras' theorem

    3. The attempt at a solution

    (Note; for ease of writing on paper, I replaced [tex]\alpha[/tex] with [tex]\theta[/tex], because, pathetic as it sounds, I don't like writing [tex]\alpha[/tex])

    A bit of a mess, one of my lines of work went;

    49sin^2[tex]\theta[/tex] + 49cos[tex]\theta[/tex] = 2401 = 49^2

    sin^2 [tex]\theta[/tex] + cos^2 [tex]\theta[/tex] = 1

    Divide all by cos^2[tex]\theta[/tex]

    (sin^2[tex]\theta[/tex])/(cos^2[tex]\theta[/tex]) + 1 = sec^2[tex]\theta[/tex]

    tan^2[tex]\theta[/tex] + 1 = sec^2[tex]\theta[/tex]

    Which, obviously, doesn't help towards my answer.

    What I'm most interested in is a kind of kick start, if I knew what kind of thing I'm supposed to do, I'd maybe be able to do the question myself, but I honestly do not know where to start.
     
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This is meant as a physics problem, not a math problem. So start with the equations for projectile motion. Write expressions for the position as a function of time, treating vertical and horizontal position separately. Combine those equations to see what you can deduce about the launch angle.
     
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