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Acceleration of gravity NOT ON EARTH

  1. Jun 11, 2012 #1
    This was the first day of class, and this is due Wed. Thank you.

    1. The problem statement, all variables and given/known data

    An object undergoes free fall 126 m on the planet whoosedat. It takes 6.00 sec. to travel the 126 m what is the acceleration of gravity on whoosedat?


    2. Relevant equations

    How long would it take an object to fall 15.0 m on whoosedat?


    3. The attempt at a solution

    1. It is distance/time? making it 21m/s

    2. 21m/s/15? making it 1.4 sec
     
  2. jcsd
  3. Jun 11, 2012 #2

    SammyS

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    Hello Justhelp. Welcome to PF !

    That gives the average velocity for the 6 seconds.

    If the object started from rest, what is its final velocity, if its average velocity is 21 m/s ?
     
  4. Jun 11, 2012 #3
    21x6=126 final velocity right?
     
  5. Jun 11, 2012 #4
    Is that what I have to do before I reach the results? I am not seeing it.

    And thanks for the welcome :)
     
  6. Jun 11, 2012 #5

    SammyS

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    No.

    For constant acceleration, [itex]\displaystyle v_{\text{Average}}=\frac{v_{\text{Iniial}}+v_{ \text{Final}}}{2}\ .[/itex] ... just an average of the two.
     
  7. Jun 11, 2012 #6
    My dad was just explaining to me per second per second.. And I couldn't quite get it. Could you explain? I understand that acceleration has to always have ms^.
     
  8. Jun 11, 2012 #7
    And I came up with 3.5 ms^.
     
  9. Jun 11, 2012 #8

    berkeman

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  10. Jun 11, 2012 #9

    berkeman

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    I'm getting a slightly different answer. Can you post your work?
     
  11. Jun 11, 2012 #10

    HallsofIvy

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    The problem tells you that the object falls 126 meters. That is a distance, not a velocity. Assuming that the distance is short enough that acceleration can be taken to be a constant, g, then the change in speed over t seconds is gt and the distance fallen is [itex](g/2)t^2[/itex] (assuming that the initial speed was 0). With t= 6 seconds, you have [itex](g/2)6^2= 126[/itex]. Solve that for g and then complete the problem.
     
  12. Jun 11, 2012 #11
    a=9.8 m/s^

    d= a (m(s^)

    126= a (m(s^)

    126=a(m(6^)
    126= a(m(36)
    note: 36 is divided and cancelled on the right.

    126/36=a
     
  13. Jun 11, 2012 #12
    Hallsofivy what is itex?
     
  14. Jun 11, 2012 #13

    SammyS

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    Why are you using g for Earth, 9.8 m/s2, all of a sudden?

    If the object starts from rest, accelerates uniformly and has an average velocity of 126m/s, then it's final velocity (at the end of 6 seconds) must have been 252 m/s.
     
  15. Jun 11, 2012 #14
    Because the problem I have come to conclude is not giving the gravity, it may be implying it but I can not tell, the implied gravity in general is 9.8/m s^
     
  16. Jun 11, 2012 #15
    Thanks for fixing that Hallsofivy :). Your solution is very much similar to mine. :) it helps!
     
  17. Jun 11, 2012 #16

    HallsofIvy

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    You keep writing "/ms^". What in the world does that mean? Speed or velocity is "distance divided by time", m/s or "meters per second". Acceleration is speed divided by seconds and so "(meters per second) per second", m/s^2.

    "the implied gravity in general is 9.8/m s^". Where is that implied? The whole point of the problem is to calculate the acceleration due to gravity from [/itex](a/2)(6^2)= 126[/itex]. Simple algebra tells you that a is NOT 9.8!
     
  18. Jun 11, 2012 #17
    I mean, meters per second sqaured.
     
  19. Jun 11, 2012 #18
    9.8 being the gravity on earth..
     
  20. Jun 11, 2012 #19

    HallsofIvy

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    Yet, the whole point of the problem is that this is NOT on the earth so that is completely irrelevant.
     
  21. Jun 11, 2012 #20
    a is the gravity # 9.8 is just a ref. I didn't use 9.8 at all in the problem.
     
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