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Helicopter problem

  1. Feb 28, 2008 #1
    1. The problem statement, all variables and given/known data
    The height of a helicopter above the ground is given by h=2.55t^3, where h is in meters and t is in seconds. At 1.85 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?


    2. Relevant equations



    3. The attempt at a solution

    _____________________________h=2.55t^3
    1.85s

    h=2.55(1.85s)^3

    Thank you very much
     
  2. jcsd
  3. Feb 29, 2008 #2
    I'm fairly sure that you can make use of the formula:

    [tex]h=v_{i}t+\frac{1}{2}at^2[/tex]

    And do it as you started:

    [tex]2.55(1.85)^3=v_{i}t+\frac{1}{2}at^2[/tex]

    And then initial velocity is zero, and a is the gravitational constant [tex]g=9.8m/s^2[/tex]:

    [tex]2.55(1.85)^3=0t+\frac{1}{2}9.8t^2[/tex]

    And then solve for t.
     
  4. Feb 29, 2008 #3
    WHO IN THE WORLD TOLD YOU THAT THE INITIAL SPEED OF THE BAG IS ZERO???
     
  5. Feb 29, 2008 #4
    Thank you very much

    If it isn't 0, how would you solve it?

    Would you do this?

    velocity:
    h=3.25t^3
    h'=9.75t^2
    h'(1.65)=9.75(1.62)^2
    =26.5444

    position:

    h=3.25(1.65)^3
    =14.6m

    I would then use all of this to solve for t, right?

    h=vit+1/2at^2
    14.6m=26.5444m/s+1/2(-9.8m/s^2)t^2
    t=1.56s

    Thank you
     
    Last edited: Feb 29, 2008
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