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Webassign - Dropping a Package From a Helicopter

  1. Sep 11, 2012 #1
    Q: A helicopter is ascending vertically with a speed of 2.00 m/s. At a height of 160 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

    Attempt:
    [how long it takes to reach the height]
    v=delta d/time
    2=160/time
    2(time)=160
    time=80 seconds

    [how long it takes to fall]
    v=delta d/time
    9.8=160/time
    9.8(time)=160
    time=16.326

    80+16.326

    96.3 s (which is wrong.)
     
  2. jcsd
  3. Sep 11, 2012 #2

    Doc Al

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    Staff: Mentor

    The time it takes for the helicopter to reach the given height is irrelevant. They just want the time it takes for the package to fall to the ground, starting from the moment it is dropped.

    9.8 m/s^2 is the acceleration of the falling package, not the speed.

    How can you express the position of the package as a function of time? It is uniformly accelerated motion. (You'll need to review your kinematic equations. Here's a list: Basic Equations of 1-D Kinematics)
     
  4. Sep 11, 2012 #3

    CWatters

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    Homework Helper

    Read the question again. When the package is released at 160m it is travelling upwards at same speed as the helicopter. It will carry on upwards for a bit. You need to work out how much higher it goes and add that to the 160m. Also calculate the time that phase takes.

    Hint: Use one of the equations of motion to account the deceleration due to gravity.

    Having worked out the height it reaches in part one you can then work out how long it takes to fall.

    Then add the time for each phase together.

    Hint: use one of the equations of motion to take into account the acceleration due to gravity.
     
  5. Sep 11, 2012 #4
    the upward motion the package will have will be rather insignificant because the 2m/s speed isn't fast enough to make a difference in the problem, keep in mind this is a basic physics homework, the person that made the exercise wasn't probably thinking of the upward motion, since this is probably a high school exercise.
     
  6. Sep 11, 2012 #5

    CWatters

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    The difference is about 7%.
     
  7. Sep 11, 2012 #6

    Doc Al

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    I suspect that the author of the problem was well aware of the upward motion--that's part of the point of the exercise. Such problems are standard fare.
     
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