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Interesting angle of elevation problem

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the angle of elevation of a projectile that travels THREE time horizontally its MAXIMUM height, on earth

    g = 9.800m/s2


    2. Relevant equations

    Equation 1 - H = ((vsinθ)2)/2g

    Equation 2 - R = 2v2((sinθcosθ)2/g


    3. The attempt at a solution

    I tried making 3x(equation 1)=equation 2, and letting v=5m/s, and then solving for θ, but I don't think I rearranged the equation properly. THE ANSWER IS FOR SURE 53.13°, I JUST HAVE TO SHOW HOW TO GET IT
    Thanks for any help
     
    Last edited: Oct 18, 2012
  2. jcsd
  3. Oct 18, 2012 #2

    lewando

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    Equation for range is not right.
     
  4. Oct 18, 2012 #3
    What is the equation for range? And could you help me with it?
     
  5. Oct 18, 2012 #4

    lewando

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    R = 2v2sinθcosθ/g. That should get you going.
     
  6. Oct 18, 2012 #5
    i still can't figure it out
     
  7. Oct 18, 2012 #6

    lewando

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    Nothing wrong with your original approach (3H = R). Show your work so you can more specifically show where you are having a problem.
     
  8. Oct 18, 2012 #7
    I dont need help with that one anymore, but if you could help me with a different problem, it would be great.

    a)To what elevation must one go to decrease your "weight" to one sixteenth (1/16) of its "surface" value?

    b) At the elevation calculated in part a), what is the period of this orbit?

    G = 6.670x10^-11
    R(earth) = 6.380x10^6
    M(earth) = 5.98x10^24
     
  9. Oct 18, 2012 #8

    lewando

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    You should start a new thread, following the template, listing all relevant equations, showing your work, etc.
     
  10. Oct 18, 2012 #9
    thats all i have, i have no idea how to solve it
     
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