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Projectile Motion given a vertical distance and angle at that point

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data
    jgi9b.jpg

    2. Relevant equations
    Famous Five equations for uniform accelerated motion

    3. The attempt at a solution
    To be honest, I have absolutely no idea where to start. The only thing that I notice is that at the point where you're given the velocity vector (when the projectile is 10m above the ground) it makes an angle of 45 degrees, which means that both vx and vy are equal. I don't really know where to go from there.
     
  2. jcsd
  3. Sep 26, 2011 #2
    I'm glad you recognize Vx and Vy are equal. That is key to being able to solve the problem. :smile:

    Since you know Vx and Vy are equal, think about what else you know.
    What is the acceleration in the x-direction?
    What about in the y-direction?

    Knowing the starting position (just clearing the wall) can you use any of those "Famous Five" with Vx and Vy and your known accelerations to set up an expression that will let you solve for more information?
     
  4. Sep 26, 2011 #3
    There is no acceleration in the x direction, delta d x = vx delta t.
    There is acceleration in the y direction, delta d = viy delta t + 1/2 g delta t squared

    delta d y = delta d x x time + 1/2 g t squared --> i end up with -18 = 4.9t^2 which can't be done.
     
  5. Sep 26, 2011 #4
    NVM! got it YAY!
     
  6. Sep 26, 2011 #5
    Alright then. Glad to hear it. :approve:
     
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