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Throw a ball, initial velocity?

  1. Dec 17, 2015 #1
    1. The problem statement, all variables and given/known data
    A referee throws a ball in a basketball game. When the ball was in his hand the heigh was 1,6 meters above the ground. The ball is caught 0.8 seconds later by a jumping player. The ball is 2.7 meters above the ground and on the way down. what was the initial velocity when the ball left the referee's hand?

    2. Relevant equations

    s=v0t + at^2/2
    v=v0 + at

    3. The attempt at a solution

    I am guessing that the basketball player caught the ball when it was at its highest point and therefor the velocity was 0 m/s.

    0 = v0 + at
    at = - V0
    -9,86 * 0,8 = - v=
    9,82 * 0,8 = v

    this answer is wrong so i tried again with the second equation and that gave the right anwer. Why is that?
  2. jcsd
  3. Dec 17, 2015 #2


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    Well the question says he caught it "on the way down", not at its highest point.

    Since you know the distance and the time, but not the initial nor final velocity, I think you used the wrong equation to start with.
  4. Dec 17, 2015 #3


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    Note that the ball is "on the way down" when it is caught. Is that compatible with your assumption that v = 0 at the time the ball is caught?
    Edit: [I see Merlin already pointed this out.]

    I think the problem should have stated that the ball was thrown vertically.
  5. Dec 17, 2015 #4


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    Ooops! My final comment is wrong. You may use that equation first or second, so long as you don't assume you know v.

    In fact I think you may need to get two equations involving v and v0 so that you can eliminate v.

    You know time and distance at both start and end, also you know the acceleration, but you have two unknown velocities.
  6. Dec 18, 2015 #5
    but now that I think about it why can I use the second equation? cuz when the ball is going up its acceleration is -9.82 but when it is on its way down the acceleration is 9,82.
  7. Dec 18, 2015 #6


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    In answer to your query in #5, the second equation , v=v0 + at has two unknowns and you get v= v0 -9.8 x 0.8 which does not get you an answer yet.

    I'm not sure what your second comment is about. gravity is a constant in this problem, so always has exactly the same value.
    If we are taking UP as positive, then a = - 9.8 for all parts of the trajectory, because it is an acceleration downwards.

    Now back to the story. First let me apologise for maybe leading you astray. It was late last night and I obviously was not thinking too clearly.
    If you just use your second equation s = v0 x t + a x t2/2 you know everything except v0 so you can solve for that.

    Write down all the values you know in the problem (so that I can check that you have these right) then attempt to solve that second equation.
  8. Dec 18, 2015 #7


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    You have to choose which direction is positive, up or down, and stick to it for all distances, velocities and accelerations.
    If up is positive then, as Merlin posted, the acceleration is negative throughout. The velocity starts as positive and finishes as negative, so on the way up the negative acceleration attenuates the positive velocity, but on the way down it augments the negative velocity.

    As TSny posted, there is not enough information. I'm not familiar with the formalities of basketball. If there is some context in which the referee would throw the ball straight up then that is probably what is missing. Otherwise, assume it meant to ask for the initial vertical velocity. Or perhaps you omitted some part of the question statement?
  9. Dec 18, 2015 #8
    It doesn't make any sense for the referee to throw the ball with a horizontal component since that would give one team an advantage.
  10. Dec 18, 2015 #9
    This is a common misconception. Note that at the highest point in the trajectory the acceleration is not zero!

    You do not need to break the analysis of the motion into two separate parts. Equations like ##\Delta x=v_ot+\frac{1}{2}at^2## will give you the displacement regardless of whether or not the object changed direction during the displacement. So, for example, if it starts at a height of 0.9 m, travels upward to a height of 1.5 m, and then falls and is at a final height of 0.8 m, ##\Delta x## will be -0.1 m. On the other hand, if it starts at a height of 0.9 m and moves downward the entire time ending at a height of 0.8 m ##\Delta x## will also be -0.1 m, and that one equation is valid in both circumstances.

    @haruspex In basketball there's a tip off, or jump ball, that occurs at the beginning of each period, or any time when two opposing players have a dispute over possession. The referee throws the ball directly upward and the two players jump, attempting to tip (or tap) the ball towards one of their team mates.
  11. Dec 19, 2015 #10


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    That's not correct. Gravity always acts in the same direction = downwards.
  12. Dec 19, 2015 #11
    What if the ball starts at 1,6 meters and falls at 2,7 meters? is the displacement 1,1 because it is higher than the start or is it -1.1 cus it is FIRST Height - Last height?= displacement
  13. Dec 20, 2015 #12
    I don't understand that statement. If the initial height is 1.6 m and the final height is 2.7 m, then the displacement is +1.1 m. We wouldn't call that a "fall"!
    Always final position minus initial position. This is the reason velocity is negative when the position is decreasing, and positive when the position is increasing.

    ##\bar{v} = \frac{\Delta x}{\Delta t}##
    Last edited: Dec 20, 2015
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