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One / Two Dimensional Motion

  1. Sep 18, 2010 #1
    3. The attempt at a solution[/b

    I tried to use the

    X= Vixt + 1/2axt^2 => 0= 1/2axt^2+ Vixt + X

    To find the time since there isn't a horizontal distance given.

    Then I realized the initial velocity for x and y direction can't be the same , but I don't have the angle.

    Do i have to find the angle first?

    If so, what do I use ?

    Edit: I understand how to solve it now. Initial velocity in y direction is zero so I can solve for the time using the same equation above.

    Here is the question

    Attached Files:

    Last edited: Sep 18, 2010
  2. jcsd
  3. Sep 18, 2010 #2
    It is a horizontal projection so the velocity at the beginning only has an x component. And only the height of the tower is given.
    hint: freefall
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