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Cannon n Hoops and deltaX

  1. Oct 5, 2007 #1
    Hello fellow forum trolls,

    I am currently pondering about this physics problem that was assigned to me on a day I was not in school and I have some confusion on the process in which to solve it.


    Essentially we have to cannon out a metal ball through a hoop a certain distance away.

    We know the initial velocity of the ball, The DeltaY of the hoop, and we have control over the DeltaX. To pass the project we must determine where to set the Hoop in respect to the launcher to get the ball through.

    so far my data goes as follows

    The launcher is at 30 degrees launching out at a velocity of 5.85

    .............X.................Y
    Vi..........5.073 m/s....2.92 m/s
    Vf..........5.073 m/s
    D...........................| .265m
    a...........0...............-9.8 m/s^2
    t


    So my questions are,

    For my final velocity of Y i got 3.711 but I feel that it doesnt seem write. I used the formula v^2=Vo^2+2a(DeltaY)

    After that I cannot figure out how to find DeltaX

    If you could point me in the write direction or help check my work that would be amazing.

    Thanks for your help
     
  2. jcsd
  3. Oct 5, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    So delta Y is 0.265m?

    It appears you took a as +9.8m/s^2 instead of -9.8m/s^2... when finding the final velocity...

    What I'd do is first find the path of the projectile in the form y = f(x)...

    You have your two displacement equations:

    x = vocos(theta)*t

    y = vosin(theta)*t - (1/2)gt^2

    try to get an equation with just x and y... then plug in the y value you need and solve for x... you'll get 2 answers.
     
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