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Kinematics - finding time given height and acceleration

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data
    The height of a helicopter above the ground is given by h = 2.50t^3 , where h is in meters and t is in seconds. After 2.45 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

    2. Relevant equations
    d=v1t +1/2at^2
    3. The attempt at a solution
    So i found the height of the helicopter at 2.45s which also = the distance the bag must fall:
    =2.5(2.15)^3
    = 36.8 m

    then since the bag is in free fall it has an acceleration of 9.8m/s^2
    and i just assume that the velocity is 0 because it is being drop from the helicopter.
    so using kinematic equation: d=v1t +1/2at^2
    rearrange for time and plugging in:
    t^2= 36.8/4.9
    t = 2.74s

    can someone just check that my reasoning is correct? that would be great thanks :)
     
  2. jcsd
  3. Sep 20, 2009 #2

    Redbelly98

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    Well, since the helicopter is moving, the bag will have a (non-zero) initial velocity. So if you can figure out the helicopter's velocity at 2.45s, then you can use that for v1.
     
  4. Sep 20, 2009 #3
    im not sure if im finding velocity correctly...
    so d = 36.76 m
    a = 9.8 m/s^2
    t = 2.45 s
    plug into kinematic equation: d= vt + 1/2at^2 = 3 m/s

    or do i not use gravity as acceleration because the helicopter could be accelerating...??
     
  5. Sep 21, 2009 #4

    Redbelly98

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    The helicopter is rising, not falling with gravity. Also, its acceleration is not constant; since a is not constant, those equations for constant acceleration do not work for the helicopter.

    Have you had calculus? The derivative would be helpful in figuring out the helicopter's velocity here.
     
  6. Sep 21, 2009 #5
    ok so the velocity of the helicopter h'= 7.5t^2 , plugging in 2.45s means the bag has in initial velocity of 45.0 m/s ...???
     
  7. Sep 21, 2009 #6

    Redbelly98

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