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Finding velocity (having trouble interpreting the question

  1. Jan 22, 2015 #1
    1. The problem statement, all variables and given/known data
    For an initial velocity of V0, what is the velocity of the ball as it passes the point A on its way up, where OA is 1/3 the maximum height.

    V0 = 10m/s
    t = 1s (2s total for the ball to come back down)
    maximum height = 5m
    initial displacement x0 = 0m

    2. Relevant equations
    Vf^2 = V0^2 + 2a(x - x0)

    3. The attempt at a solution
    6jO1eUG.png

    I'm assuming point O is the ground. If point OA is 1/3 of the maximum height then, it is 5/3m. My confusion is what point A should be. Is it some arbitrary point that is just deemed "A"? The answers are in terms of vA (velocity as the ball passes through point A) = initial velocity times a constant. The other thing I can think of is that I am wrong for setting point O at 0. Am I supposed to go another "O" distance from OA to get to point A?

    I know this answer is yahoo answers or something, but I refrained from just looking at the answer because I really want to understand what is going on.
     
  2. jcsd
  3. Jan 22, 2015 #2

    Bystander

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    You're shown a parabolic path/trajectory. "O" is at ground level, or y = 0 describes a line "O" which you could regard as the "x-axis."
    Think "y = A."
    Call it an "arbitrary" altitude --- "A" for "arbitrary" or for "altitude" or for "ambiguous" statement.
     
  4. Jan 22, 2015 #3
    Correct.
    No. According to the information you gave it is at x = 5/3 m.

    Is this supposed to be solved algebraically, or is the answer supposed to be for the specific problem of v0 = 10m/s, max ht = 5 m?

    No. These are both correct interpretations.

    If you are trying to do this algebraically, then you use your equation to determine the maximum height x in terms of v0. To do that, you need to select the gravitational value of the acceleration a.

    Chet
     
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