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1d Kinematics Homework: Helicopter mailbag

  1. Sep 24, 2014 #1
    1. The problem statement, all variables and given/known data
    The height of a helicopter above the ground is given by h = 3.30t3, 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.

    Already solved for Initial velocity for when mailbag's released, V0=32.08 m/s
    and also solved for initial height at 19.2 m and maximum height at 71.8 m.
    Acceleration of gravity = -9.81 m/s/s

    The actual question I have is What is the velocity of the mailbag when it hits the ground?


    2. Relevant equations
    Vxf2 = Vxi2 + 2(ax)(xf - xi)

    3. The attempt at a solution
    I plugged in 32.08 = Vxi , Ax = 9.81 and delta x(xf-xi) = 52.6 and I got 45.39 after taking the square root, which I know is wrong because I got the real answer by accident, the real answer being 37.96 m/s. I got the real answer by not squaring 32.08 and having delta x being 71.8. So basically this: Vxf = sqrt(2(9.81)(71.8)+32.08)

    When I was looking over at YahooAnswers, I found that the delta x was something like -19.2, and acceleration is of course -9.81, so I plugged that in and got 37.49 m/s, which is extremely similar to my accident answer.

    So basically my query is this: was my delta x wrong? Was I supposed to subtract 0 and 19.2 and use that as my delta x? Just want to know how to do this type of problem correctly instead of accidentally stumbling on it.
     
  2. jcsd
  3. Sep 24, 2014 #2
    Where did get that 52.6 value anyway?
    And what do you call "maximum height"? Is the text of the problem given completely here?.

    It's not a good idea to just randomly try various values.
    What is the height of the helicopter when it drops the bag? What is the final height reached by the bag? What is the difference?
     
  4. Sep 24, 2014 #3
    52.6 is delta x. The Final height (71.8) minus the initial height (19.2).

    There were two prior sub-questions for this question.
    A) What is the velocity of the mailbag when it is released? Which is obviously 32.08 m/s
    B) What maximum height from the ground does the mailbag reach? Which is 71.8 m. And I solved this using Xf = 19.2 m [This value I got from plugging in 1.80 into the height function] + (32.08)(3.27) - (1/2)(9.81)(3.27)2

    Sub-question C is the one I posted in the original post.

    In any case. To answer your question of "What is the difference", I wouldn't know, since I'm in introductory physics. And I didn't plug in just any random values. I plugged in what I thought to be delta x at the time.
     
  5. Sep 24, 2014 #4
    Unless there is more text to the problem which you did not post here, these heights make no sense to me.
    Why would be a "maximum height"? That formula for height does not produce a maximum.

    It says that the mailbag is released "after 1.80 s". Is this time measured from t=0?
    If this is the case, at what height is the helicopter when the mailbag is released?

    Regarding the final height, where will the bag stop when dropped from helicopter? At what height? You don't need advanced physics for this,
     
  6. Sep 24, 2014 #5

    NascentOxygen

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    Staff: Mentor

    There are 2 ways to work out its velocity just as it encounters the ground. You could consider it falling (initial velocity =0) under gravity from height 71.8m.

    Alternatively, recognize that if it is released with some speed upwards, it will have regained that precise speed by the time it returns to the height it was released from. So you could consider it travelling 19.25m downwards under gravity, but starting with a velocity of 32.08m/s downwards.

    Either method should work.

    The scenario would be more realistic to involve maybe a rocket, as the downdraught from the rotor of a helicopterwould wreak havoc on a mailbag's trajectory!
     
    Last edited: Sep 25, 2014
  7. Sep 24, 2014 #6
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