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Homework Help: Projectile Motion find the highest angle

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data

    http://postimg.org/image/7yfw1zpq1/ [Broken]

    http://postimg.org/image/7yfw1zpq1/ [Broken] <-Question

    3. The attempt at a solution
    I dont really understand the question but ill try.

    I need to first find the highest angle at which the projectile will clear the peak.
    At Peak:

    Sy = 0.5(u+Vpeak)t , vpeak = 0
    1800 = 125tsinx
    tsinx = 14.4
    t = 14.4/sinx -equation1

    Sx = 0.5+(Ux+Vpeakx)t , Ux=Vx i guess?
    2500 = 0.5(250cosx)t
    2500 = 125tcosx
    tcosx = 20
    t = 20/cosx
    Substitute eq 2 into eq 1:
    20/cosx = 14.4/sinx
    cosx = 1.39sinx

    Then i dunno how to solve
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 6, 2014 #2

    Orodruin

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    You need to rethink a bit. It is not necessary that the vertical velocity is zero when passing the peak. Try finding the height as a function of the horizontal distance for an arbitrary angle.
     
  4. Sep 6, 2014 #3
    need more tips lol
     
  5. Sep 6, 2014 #4
    This is a nice problem. I agree with Orodruin, that is probably the easiest method. What is your difficulty in following it?
     
  6. Sep 6, 2014 #5
    At top of parabola Vy = 0

    Vy = Uy + at
    0 = 250sinx + 9.81t
    t = 25.5sinx

    After that?
     
  7. Sep 6, 2014 #6

    Orodruin

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    Yes, but the top of the parabola is not necessarily where the projectile passes the mountain top.
     
  8. Sep 6, 2014 #7

    SteamKing

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    The top of the trajectory of the projectile can be reached before the peak is encountered, at the peak, or after the peak is encountered, so long as the altitude of the projectile clears the top of the peak.
     
  9. Sep 6, 2014 #8

    Orodruin

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    How does this contradict what I just said?
     
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