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A ball shot at an angle

  1. Oct 27, 2004 #1
    If a ball is shot out at 5.4m/s and makes an angle of 30 degrees with the horizontal, the x component would be 5.4cos30 and the y component would be 5.4sin30 correct?

    The question then asks me to write an expression for the ball’s velocity, v, using unit vectors for the x-direction and the y-direction. Wouldn’t the velocity be just 5.4m/s? I don’t understand wht they are asking.
  2. jcsd
  3. Oct 27, 2004 #2
    Unit vectors are a simple way of describing a 3d (or 2d) vector. Basically, you put everything that describes the x-direction together and put 'ex' with a vector sign over it behind it. Do thesame for the y-direction.
    In this case: the x-component of the speed is just 5.4cos(30), so that part of the vector is (5.4cos(30))*ex (actually make that x subscript and add an arrow above the e)

    The y-component is thesame thing, only with 5.4sin(30) and ey. In case they ask you to do the expression for t, you'll have to add acceleration like this:
    V = (5.4cos(30))*ex + (5.4sin(30) + ..acceleration*t..)*ey
    This describes the vector.
  4. Oct 27, 2004 #3
    What if the question wanted me to sketch the vector and find the magnitude and direction of motion with respects to the x-axis?
  5. Nov 2, 2004 #4
    Sorry for the late reply:
    Just sketch an arrow making 30 degrees with the horizontal. Then project the vector onto the x-axis and that's your x-component. It's that easy.
    Magnitude of the vector in the x-direction is as you described, 5.4cos30. The direction of motion with respect to the x-axis will be 30 degrees.
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