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Homework Help: Dynamics in Two Dimensions Question

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data

    During baseball practice, you go up into the
    bleachers to retrieve a ball. You throw the
    ball back into the playing field at an angle of
    42° above the horizontal, giving it an initial
    velocity of 15 m/s. If the ball is 5.3 m above
    the level of the playing field when you throw
    it, what will be the velocity of the ball when
    it hits the ground of the playing field?


    2. Relevant equations



    3. The attempt at a solution

    5.3m=-0.5(9.8)t^2+15(sin 42°)Δt

    vf=(15)(sin 42°)+(-9.8)(1.02)

    vx^2 + vy^2 (square root)
    (15 x cos 42)^2 + 0.041^2

    what am i doing wrong? Answer is 18m/s.
    Last edited: Aug 20, 2013
  2. jcsd
  3. Aug 20, 2013 #2


    User Avatar
    Homework Helper
    Gold Member

    Your problem is localized here:
    The result you got for t is not correct.
  4. Aug 20, 2013 #3
    This looks more like projectile motion. In this case, your equation: 5.3m=-0.5(9.8)t^2+15(sin 42°)Δt is not necessary.

    It seems to be more useful as [itex]Y=Y_{o}+V_{oy}t-\frac{gt^{2}}{2}[/itex].
  5. Aug 21, 2013 #4


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    Science Advisor
    Homework Helper
    Gold Member

    It starts at 5.3m above the final height, so what is the change in height?
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