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Relative Velocity along a Straight Line problem

  1. Jul 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A helicopter 8.50 m above the ground and descending at 3.50 m/s drops a package from rest (relative to the helicopter). Just as it hits the ground, find a) the velocity of the package relative to the helicopter and b) the velocity of the helicopter relative to the package. The package falls freely.

    2. Relevant equations: package: p , helicopter:h, ground:g
    Vp/h = Vp/g + Vg/h (I think this equation might help, but I was unable to successfully solve the problem).

    3. Answer: a) 9.9 m/s downward b) 9.9 m/s upward

    Thank you! I greatly appreciate your help in solving this problem! :)
     
  2. jcsd
  3. Jul 10, 2009 #2

    turin

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    I don't know what that equation means.

    Are you familiar with "the kinematic equations"?
     
  4. Jul 10, 2009 #3
    Yes I am, but I don't know what equation to apply to this problem to solve it. The equation I gave was listed in my textbook as a way to solve relative velocity problems:

    velocity of the package relative to the helicopter = (velocity of the package relative to the ground) + (velocity of the ground relative to the helicopter)

    Note: velocity of the ground refers to velocity of the earth
     
  5. Jul 10, 2009 #4

    turin

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    Ah, OK, now I know what your equation means. You will need it, but you will also need (at least) one more kinematical equation. Think about what variables in the equations you know and which ones you're trying to find. Also make sure that the equation that you use is applicable to the situation (e.g. if the acceleration is nonzero, then discard all equations that assume constant velocity).
     
  6. Jul 10, 2009 #5
    Thank you so much! I will try my best to solve the problem and let you know what I come up with
     
  7. Jul 10, 2009 #6
    My answer to part a)

    Ok, I used the kinematical equation Vy^2 = Voy^2 + 2ay(y-yo) to find the velocity of the package relative to the ground and I got 12.9 m/s downward (or -12.9m/s). I plugged this value into my equation as Vp/g. I know that the velocity of the helicopter relative to the ground (Vh/g) is 3.50m/s downward (or -3.50 m/s). So Vg/h, which equals -Vh/g, is +3.50 m/s. I then solved for Vp/h, and got 9.4 m/s downward...not quite 9.9m/s...did I solve the problem correctly?
     
  8. Jul 10, 2009 #7

    turin

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    Looks OK. I didn't check the numbers.
     
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