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Ball thrown upward

  1. Sep 19, 2006 #1
    A ball is thrown upward from the edge of a building that is 100m tall. On its way back down, the ball just misses the building, falling to the ground below. The elapsed time between the throw and the ball hitting the ground is 6.5 sec.


    A) The initial speed of the ball
    B) Position and speed of the ball after 1 sec and after 4 secs.
    C) The maximum height reached by the ball, measured above the ground.

    *Ok, I've worked a few problems where the object is dropped directly from the top of the building but none where the object is actually thrown upward!*
     
  2. jcsd
  3. Sep 19, 2006 #2

    radou

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    Hint: initial position.
     
  4. Sep 19, 2006 #3
    so...

    So, would the initial speed be 0m/s or is that only when the ball is at rest?
     
  5. Sep 19, 2006 #4

    radou

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    No, the initial speed does not equal zero. Remember, the ball is thrown. You can get the initial speed directly from the equation which represents the displacement of the ball, x(t) = x0 + v0*t + 0.5*g*t^2. Just set the x-axis along the building and everything will be more clear.
     
    Last edited: Sep 19, 2006
  6. Sep 19, 2006 #5
    Just x0, not x0*t.
     
  7. Sep 19, 2006 #6

    radou

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    Thanx, I mistyped. :)
     
  8. Sep 19, 2006 #7

    when you say, (t) = x0*t + v0*t + 0.5*g*t^2 does that translate to
    initial displacement times time, plus initial velocity times time, plus 0.5 times gravity times time squared?

    would the initial displacement equal 100 meters? time 6.5 sec?
    so, I just plug and chug.....
     
  9. Sep 19, 2006 #8

    radou

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    Once again, I apologise for my mistype. x(t) = x0 + v0*t + 0.5*g*t^2. Yes, that translates to initial displacement (100 m) plus velocity times time etc. The time is 6.5 seconds. BUT, there is one more condition. The left side of the equation has to equal zero, since at that time the ball fell on the ground, and we have set the coordinate axis x so that the origin (x=0) is at the bottom of the building (the ground), and x = 100 m is the point from which we threw the ball.
     
  10. Sep 19, 2006 #9
    so, 0= 100m + v0*6.5s + 0.5*-9.80m/s^2*6.5s^2 ? then I solve for the initial velocity?
     
  11. Sep 19, 2006 #10

    radou

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    Yes, exactly. I'm glad you noticed g must be negative.
     
  12. Sep 19, 2006 #11
    hmmm


    ok, I have 0=100m + v0 * 6.5s + 0.5 * -9.80m * 6.5 after I cancel out the s^2. I'm not sure how I add the meters and seconds..... shouldn't I have seconds in the demoninator? Would I use pemdas rule to add these?
     
  13. Sep 19, 2006 #12

    radou

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    What? You don't have to add any meters nor seconds. You're dealing with numbers only.
     
  14. Sep 19, 2006 #13
    I have 100 m and 6.5 s. I'm confused now...
     
  15. Sep 19, 2006 #14

    radou

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    Yes, and you have all you need. You already wrote down the equation you need to solve. Again, 0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2 .
     
  16. Sep 19, 2006 #15
    So, how should I go about multiplying and adding the numbers? I tried to multiply and add the numbers out but i got -106.525, which can't be the answer...
     
  17. Sep 19, 2006 #16
    i think i have it


    0 = 100 + v0*6.5 + 0.5*(-9.81)*6.5^2
    = 100 + v0*6.5 + (-207.025)
    = -107.025 + v0*6.5
    = 107.025= v0*6.5
    = 16.5= v0
    so, 16.5 m/s =v0 ?
     
  18. Sep 19, 2006 #17

    radou

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    Yes, looks good. I'm sure you could do this by yourself (the last step), and I don't think such steps need to be discussed here. :smile:
     
  19. Sep 19, 2006 #18
    where did you..


    where did you get the formula from? where does the 0.5 come from in the formula?
     
  20. Sep 19, 2006 #19

    radou

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    These are basic kinematic equations which can be found everywhere, and it would be weird of your teacher to assign problems to you for which you do not posess the knowledge to solve; I believe he mentioned these equations at some point. :wink:
     
  21. Sep 19, 2006 #20
    thanks

    I couldn't find it anywhere in my book. I still don't understand where the 0.5 came from. I really want to know how physics works instead of plugging and chugging. Thanks for your help. Without brainiacs like you, I would be so lost..
     
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