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Helicopter and Kinematics Problem

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The height of a helicopter above the ground is given by h=2.90t^3, where h is in meters and t is in seconds. After 1.80 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative.

    A) what is the velocity of the mailbag when it is released?
    B) What max height from the ground does the mailbag reach?
    C) What is the velocity of the mailbag when it hits the ground?
    D) How long after its release does the mailbag reach the ground?


    2. Relevant equations

    Kinematic equations



    3. The attempt at a solution

    So for the first part shouldn't the velocity at release be zero? if not what time interval would I use to derive velocity at release?

    Also, max height should be 2.90(1.80)^3 = 16.9 m

    if part a is zero then the final velocity can be derived by equation Vf^2 - Vi^2 = 2a * deltaX.
    = -18.2 m/s

    the velocity

    However, the online service won't accept my answer. :/
     
  2. jcsd
  3. Sep 22, 2009 #2
    would the velocity be zero? the helicopter is moving up, right?

    it gives you the formula;

    h = 2.6 t^3 which you could also say x = 2.6 t^3

    does this look familiar? think about differentiation.

    what's the change of displacement with respect to time? (in this case) ([tex]\frac{dx}{dt}[/tex])

    from that you can work out it's velocity at release.

    just a hint but you might also think, maybe the helicopter is accelerating upwards? it's not a very linear relationship is it? :eek: so try to think about that aswell

    (hint: [tex]\frac{d^2 x}{dt^2}[/tex])
     
  4. Sep 22, 2009 #3
    No, velocity at release would be equal to the velocity of the helicopter at the time of release.
    [tex]v = \frac{dh}{dt}[/tex]
     
  5. Sep 22, 2009 #4
    2.9t^3 dh/dt = 5.8t^2

    V-heli at 1.80 sec = 18.8 m/s = Vi-package which Webassign will not accept.

    please help and thanks for the quick reply!
     
  6. Sep 22, 2009 #5
    2.9 * 3 isn't 5.8!! =P
     
  7. Sep 22, 2009 #6
    Oh i knew that....lol sorry I was in a rush to go to class
     
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