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Find the initial velocity of the package relative to the ground

  1. Sep 22, 2016 #1
    1. The problem statement, all variables and given/known data
    A helicopter is flying in a straight line over a level field at a constant speed of 6.8 m/s and at a constant altitude of 9.4 m. A package is ejected horizontally from the helicopter with an initial velocity of 12.0 m/s relative to the helicopter, and in a direction opposite the helicopter's motion. Find the initial velocity of the package relative to the ground.

    What is the horizontal distance between the helicopter and the package at the instant the package strikes the ground?

    What angle does the velocity vector of the package make with the ground at the instant before impact, as seen from the ground?

    2. Relevant equations
    Initial Velocity:
    Vhg = 6.8 m/s
    Vph = -12.0 m/s
    Vpg = -5.20 m/s
    Y = 9.8 m
    3. The attempt at a solution
    I was thinking that I could use a kinematic equation to solve for this question but I think it is much simpler than that.
     
  2. jcsd
  3. Sep 22, 2016 #2

    gneill

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    Did you have a question? You haven't given us anything we can examine or comment on in terms of a solution attempt. Can you show in detail what you've tried?
     
  4. Sep 22, 2016 #3

    kuruman

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    It is. How did you get Vpg which is what you are looking for?
     
  5. Sep 22, 2016 #4
    To find Vpg I set it equal to Vhg + Vph. I am not sure why I did that. I only know that the package is going opposite of the helicopter so its velocity is negative.
    Vpg = Vhg +Vph
     
  6. Sep 22, 2016 #5

    kuruman

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    Are you familiar with the double-subscript convention for adding velocities? Most textbooks have it. Check yours.
     
  7. Sep 22, 2016 #6
    To find the horizontal distance I tried Pythagorean theorem: x^2 + 9.8^ = -12.0 ^2 and I got 6.8
     
  8. Sep 22, 2016 #7
    Would that help with finding the horizontal displacement?
     
  9. Sep 22, 2016 #8
    would it be Xhp = Xhg + Xgp?
     
  10. Sep 22, 2016 #9

    kuruman

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    It would because it gives you the initial velocity relative to the ground. However, it looks like you found that and your question is about the horizontal distance
    To answer that, you cannot use the Pythagorean theorem because the path is a parabola not a straight line. You need to consider the kinematic equations as modified for projectile motion.
     
  11. Sep 22, 2016 #10
    If that's the case usually I like to set my values into x and y components:
    Y:
    (Package)
    Y = 9.8m
    Vi = -5.20 m/s
    Vf = -12.83 m/s solved by kinematic equation V^2 = Vi^2 + 2s(Y - Yi)
    g= -9.8 m/s^2
    t = 0.78s ~ V = Vi + at
    X:
    V = 6.8 m/s helicopter
     
  12. Sep 22, 2016 #11
    Ok so from my time I am multiplying it to the helicopters velocity because it is a constant and does not change, which means acceleration is 0, and end up getting 7.58 m for x but it is wrong.
     
  13. Sep 22, 2016 #12

    kuruman

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    If the helicopter is traveling at constant altitude, what is the package's initial vertical velocity?
     
  14. Sep 22, 2016 #13
    Is it 0 m/s? because it is not moving yet.
     
  15. Sep 22, 2016 #14

    kuruman

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    Correct. So in the vertical direction you have a package that starts with zero velocity and drops 9.8 m. Can you find how much time it takes to do that?
     
  16. Sep 22, 2016 #15
    yes it takes 1.41s from V = Vi +at. V is now -13.86 m/s because the initial Velocity was 0 m/s
     
  17. Sep 22, 2016 #16

    kuruman

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    Which V is this? If it is a component, it needs a subscript x or y. Is it a component?
     
  18. Sep 22, 2016 #17
    Its the final velocity of the package so Vy
     
  19. Sep 22, 2016 #18

    kuruman

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    Correct. Add the subscript so that you don't get confused.
    Now you have enough information to answer the remaining two questions. Do you see how?
     
  20. Sep 22, 2016 #19
    Yes. Thank you for your help. Sorry for the late response I went to a research meeting.
     
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