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How fast was the ball thrown?

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A boy throws a baseball directly upward. He catches it 3.2 seconds later.

    2. Relevant equations

    a=Δv/Δt

    3. The attempt at a solution

    Letting up be positive, I think the acceleration would be positive 9.8 m/s^2. I think I am trying to find the final velocity. I think that the inital velocity would be 0, right before he throws the ball. Could I go about this problem by using a=Δv/Δt to find velocity? This method seems to yield an answer of 3.0625 m/s. Thanks in advance for a helpful reply!
     
    Last edited: Nov 18, 2013
  2. jcsd
  3. Nov 18, 2013 #2

    berkeman

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    Staff: Mentor

    If up is positive, gravity's acceleration is downward, so it is negative.

    What is the equation for the vertical position of the ball versus time? Just solve for Vi given that the ball returns to the same y position after 3.2 seconds.
     
  4. Nov 18, 2013 #3
    Thanks berkeman! Are you saying that I should use this equation, but take into account that the acceleration is negative?
     
  5. Nov 18, 2013 #4

    berkeman

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    Which what where equation?
     
  6. Nov 18, 2013 #5
    Sorry, wrong demonstrative pronoun. *That* equation.

    Which is *this* equation:a=Δv/Δt.
     
  7. Nov 18, 2013 #6

    berkeman

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    Oh, now we're using BIG words! :smile:

    No, I mentioned the equation for the vertical position of the ball as a function of time. Can you start with that equation?
     
  8. Nov 18, 2013 #7

    HallsofIvy

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    Try using s(t)= -at^2+ vt+ h where a= 9.8 m/s^2, v is the initial velocity and h the initial height. Since you are talking about a ball going up and back down, you can take the initial height to be 0. The quadratic equation, -9.8t^2+ vt=0, has two solutions. One is obviously 0, the time the ball leaves your hand. The other solution is the time the ball returns to your hand which you are told 3.2 seconds. So solve -9.8(3.2)^2+ v(3.2)= 0 for v.
     
  9. Nov 18, 2013 #8
    Haha! You should stick around. :smile:

    I may be able to, but I haven't gotten into vertical position yet, unless you're talking about the change in height.
     
  10. Nov 18, 2013 #9
    Thank you!!! I will try this and post my answer in a moment.
     
  11. Nov 18, 2013 #10
    983.4496+v(3.2)=0

    Do I take the square root of both sides now? Or do I use the quadriatic equation?
     
  12. Nov 18, 2013 #11
    I think I can use the quadriatic equation, but I'm having some trouble figuring out which numbers to assign to the letters.
     
  13. Nov 18, 2013 #12

    berkeman

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    Please show more of your work. I know that Halls basically solved the problem for you, but you need to do some of this schoolwork on your own.

    Show us the definition of the Quadratic Equation. Show us how it works (look at wikipedia, for example). Then show us how you think you should use it to solve this problem.
     
  14. Nov 18, 2013 #13
    Well, I wouldn't have had trouble with that part anyway, I supposse, it's just assigning the letters that seems difficult. Here's the definition I'm looking at: http://www.purplemath.com/modules/quadform.htm


    -9.8(3.2)^2+ v(3.2)= 0

    Would -9.8 be a, v would be b, and 3.2 would be c?

    Thanks!
     
  15. Nov 18, 2013 #14
    Or would -9.8(3.2) be a, v would be b, and 3.2 would be c?
     
  16. Nov 18, 2013 #15

    berkeman

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    Actually, since the equation does not have a constant term (since the ball returns to the same height), you can just solve the equation directly for Vi. Sorry that I missed seeing that earlier. What answer do you get?
     
  17. Nov 18, 2013 #16
    I'm not quite sure yet. Do I take this:-9.8(3.2)^2+ v(3.2)= 0 and divide by v? Or subtract v?
     
  18. Nov 18, 2013 #17

    berkeman

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    Start with that equation and take the first term on the lefthand side (LHS) over to the RHS...
     
  19. Nov 18, 2013 #18
    Okay, thanks! I've done this plently of times with binomials and trinomials, but the quadriatics seem a bit harder.
    -9.8(3.2)^2+ v(3.2)= 0
    v(3.2)+9.8(3.2)=0

    Like that?
     
  20. Nov 18, 2013 #19

    berkeman

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    I'm not following what you did. My suggestion was to take the first term on the LHS over to the RHS. That would put the = sign between the two terms. Then simplify...
     
  21. Nov 18, 2013 #20
    I added the -9.8(3.2)^2 to v(3.2). Oops.

    -9.8(3.2)^2+ v(3.2)= 0
    v(3.2)=9.8(3.2)

    More like that?
     
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