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Equation help

  1. Sep 6, 2005 #1
    A ball is droped from a stadium.It hits the ground 2.29 seconds later.How high is the stadium?Do I use this equation?.5at^2(Thant is wrong isnt it?) How fast is the ball going when it hits the ground?(what equation do I use on this one?)
     
    Last edited: Sep 6, 2005
  2. jcsd
  3. Sep 6, 2005 #2

    TD

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    I'm assuming no initial speed, then the height h is given by:

    [tex]h = h_0 - \frac{{gt^2 }}{2}[/tex]

    Here, [itex]h_0[/itex] is the initial height, so what you are looking for. You choose h = 0, because that's where it hits the ground. Then fill in t and g and solve for [itex]h_0[/itex].
     
  4. Sep 6, 2005 #3
    In this case would the answer be about 25.7 meters? And How do I tell How fast the marble is going?
     
  5. Sep 6, 2005 #4

    TD

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    That seems to be correct yes.

    For the other question, use a relation between acceleration, speed and time. If time is in s and acceleration in m/s², what would give speed (m/s)?
     
  6. Sep 6, 2005 #5
    would that equation be the y='s one?
     
  7. Sep 6, 2005 #6

    TD

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    I was thinking about v = at
     
  8. Sep 6, 2005 #7
    I have another question.If something is dropped and hits the ground one second later how high is it? With that equation I got 7.1.Why is it not 9.8 meters?
     
  9. Sep 6, 2005 #8

    TD

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    Because it's not the speed which is 9.8m/s but the acceleration which is 9.8m/s².
    Are you sure you got 7.1 though?
     
  10. Sep 6, 2005 #9
    No I got 4.9 sorry
     
  11. Sep 6, 2005 #10
    Would the ball be going 22m/s before it hit the ground?
     
  12. Sep 6, 2005 #11

    TD

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    That is correct.

    You see, an acceleration of 9.8m/s² means that after a full second, the speed has increased 9.8m/s. So when dropping something with no initial speed, it only reaches the speed of 9.8m/s after the full second, so when it hits the ground in your example.

    The avarage speed was 9.8/2 = 4.9, exactly what you found :smile:

    That seems correct, approximately.
     
  13. Sep 6, 2005 #12
    Thank You so much TD
     
  14. Sep 6, 2005 #13

    TD

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    No problem :smile:
     
  15. Sep 6, 2005 #14
    Say someone threw the ball up and it didnt hit the ground until 3.53 seconds later how do I find out the ending velocity?What if it was thrown down and hit the ground 1.81 seconds later?
     
  16. Sep 6, 2005 #15

    TD

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    Thrown up would require the initial height and thrown down the initial speed, unless there is none.

    Perhaps someone else can help, I'm logging off. 2.50 AM here, good luck!
     
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