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Finding Angle of Trajectory

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A projectile is launched from the earth’s surface at initial speed vi at angle θ0 with the
    horizontal. When the projectile is at its maximum height h, it has half the speed it had
    when it was at half its maximum height h/2

    At what angle was the projectile launched?

    2. Relevant equations



    3. The attempt at a solution

    I know that the velocity vector at the highest point is vicosθ, and I know that the maximum height of the parabola is (visinθ)2 / 2g. At this point, I don't really know where to go. I'm not sure how to relate height and velocity together with what I know. I was doing this earlier and I tried to go further, but I just ended up with a lot of variable/trig function couples that don't make sense. I could take a picture of all the work I've done, but you wouldn't be able to follow it. I barely know what I tried.

    You don't really have to answer it for me, just what should the next step be?
     
  2. jcsd
  3. Sep 10, 2013 #2

    tiny-tim

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    Hi Lamebert! Welcome to PF! :smile:

    You need to translate …
    … into one of the standard constant acceleration equations :wink:
     
  4. Sep 10, 2013 #3
    Thanks for the welcome, and can you go a little further? I already knew I was going to need those equations, I just don't know how they fit together, how velocity and height at each point should relate inside of the equations.
     
  5. Sep 10, 2013 #4

    tiny-tim

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    well, you know the velocity at two different points, and the distance between them (and the acceleration),

    sooo the equation to use is … ? :smile:
     
  6. Sep 10, 2013 #5
    vf2 = vi2 + 2ad

    The problem is finding the angle theta, though. I know that vicosθ = v, right? But if I set that equal to what I solved for from the above equation, that is, v = √(gh /3), i end up with two variables again, which doesn't really help me.
     
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