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Gull shell acceleration physics problem

  1. Jun 6, 2009 #1
    A glaucous-winged gull, ascending straight upward at 5.40 m/s , drops a shell when it is 13.5m above the ground. What is the magnitude of the shell's acceleration just after it is released?
     
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  3. Jun 6, 2009 #2

    Pengwuino

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    Re: Acceleration

    What do you know about the Earth's gravity and how it acts upon objects in free fall (which this scenario is even though it may not immediately seem like it is)?
     
  4. Jun 6, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi wadini! Welcome to PF! :smile:

    When in doubt, always use good ol' Newton's second law …

    in this case, ask yourself what forces there are on the shell just after it is released …

    and so what is the acceleration? :wink:
     
  5. Jun 6, 2009 #4
    Re: Acceleration

    okay I tried finding the time by doing

    x=1/2*g*t^2
    so -13.5 m/s= 1/2(-9.81)t^2
    and I got t= 1.658
    and then I plugged that into 1/2(a)(1.658)^2
    and got acceleration from that but that is not the correct answer.... am I right for finding time? Is there an equation to find acceleration without time ?
     
  6. Jun 6, 2009 #5

    dlgoff

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    Re: Acceleration

    At 5.40 meters/second, how long will it take to get to 13.5 meters? Look at the units to see what operation you need to do to get seconds.
     
  7. Jun 6, 2009 #6
    Re: Acceleration

    Oh! so I have to cancel out the meters and once I do that I get 5.40 m's/13.5m and then t=.4 so do I just plug that into the equation 13.5 m/s = .5*a*(.4^2) ?
     
  8. Jun 7, 2009 #7

    Cyosis

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    Re: Acceleration

    The formula [itex]v=0.5 a t^2[/itex] does not exist. You're also complicating things. The gull is ascending with 5.40 m/s so what is the accelerating the shell experiences from the gull? What is the force? Which other force acts upon the shell?
     
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