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Helicopter final velocity help

  1. Jan 19, 2005 #1
    Okay...I am think that I am totally lost on this problem.

    A helicopter holding a 70-kilogram package suspended from a rope 5.0 meters long accelerates upward at a rate of 5.2 m/s^s. When the upward velocity of the helicopter is 30 meters per second, the rope is cut and the helicopter continues to accelerate upward at 5.2 m/s^2. Determine the distance between the helicopter and the package 2.0 seconds after the rope is cut.

    What I did to solve it was, I solved for final velocity, using the formula Vf=Vi+a*t and I got 40.4 m/s.
    Then I solved for the displacement by using the formula 1/2(Vi+Vf)t and I got 70.4 m.
    Is any part of this right? If not, can someone point me in the right direction please!
     
  2. jcsd
  3. Jan 19, 2005 #2

    Galileo

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    What you've calculated is the distance the helicopter has risen from the point where the rope was cut.
    You are asked for the distance between the package and the helicopter.
    So if you calculate the distance from package to the point where the rope was cut, then you can calculate the distance between the package and the copter.
     
  4. Jan 19, 2005 #3
    So, to calculate the distance between the helicopter and the package after the rope was cut, would I use the final velocity that I calculated as my initial velocity of the package, use the acceleration of gravity?
     
  5. Jan 19, 2005 #4

    Galileo

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    No, the initial velocity of the package is the velocity when the rope was cut: 30 m/s.
    So you are given that the initial velocity is 30 m/s upwards. The acceleration is downwards. So you can use the formula for the displacement under constant acceleration.
     
  6. Jan 19, 2005 #5
    Ok, after I solved it, I got an answer of 79.62 meters. Is that somewhat right, or did I screw up again? It seems like solving the problem was too easy, or am I making it more complicated?
     
  7. Jan 19, 2005 #6

    Galileo

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    You used: [itex]vt+1/2gt^2[/itex], for the package right? (Since that gives 79.62 m)
    You know it can't be right, since it would mean the package is higher than the helicopter. (which was at 70.4 m)

    Notice that the package got an initial velocity upwards, but gravity accelerates it DOWNwards. Therefore the acceleration is negative.
     
  8. Jan 19, 2005 #7
    Instead of adding, which is what I did, I would just have to subtract, which would give me an answer of 40.38. Is that a reasonable answer, I'm still confused with the velocity though. I understand gravity being negative because it is downward, but when using 30m/s as the velocity, wouldn't the package have a different final velocity, which would changed the displacement?
     
  9. Jan 19, 2005 #8
    Ok, lets start from the beginning. When the rope is cut, both the package and the helicopter have a velocity of 30 m/s UPWARDS. However, after that point, the helicopter continues to accelerate UPWARDS, whereas the package accelerates DOWNWARDS. Now, in order to find the distance between the two, the best method would be to find the distance they are from the original point where they are cut. You can use a simple displacement equation for both the package and the helicopter. But remember to the sign for the velocity and acceleration! Then you can simply subtract the distances to find the distance in between.
     
  10. Jan 19, 2005 #9
    When you say subtract the distances to find the distance in between, are you talking about subtracting the 40.38 meters from the 70.4 meters? Now I have everything confused. Really confused. Sorry if I am making this harder than it is, but I really need to understand this for my exam tomorrow. So if you wouldn't mind explaining with a lil more detail, I'd really appreciate it!
     
  11. Jan 19, 2005 #10
    Since the gravitational acceleration is downwards, it's -g, instead of +g. After you find both distances, all you have to do is substract them together.

    Displacement of the helicopter.

    Let
    x_0 = 5
    v_0 = 30
    t = 2
    a = 5.2

    [tex]
    x_1 = x_0 + v_0t + \frac{1}{2}at^2 = 75.4 m
    [/tex]

    Displacement of the box.

    Let
    x_0 = 0
    v_0 = 30
    t = 2
    a = -9.8

    [tex]
    x_2 = v_0t + \frac{1}{2}at^2 = 40.4
    [/tex]

    So,

    [tex]
    x_{tot} = x_1 - x_2 = 35
    [/tex]

    if there are any mistakes, feel free to comment.

    i am not sure whether the v_0 of the crate should be = 30, because when the rope is cut, wouldn't it be released at 0 velocity?
     
    Last edited by a moderator: Jan 19, 2005
  12. Jan 19, 2005 #11
    Your solution looks good to me.

    It would be thirty also. Think of conservation of momentum. The initial and final momentum must be equal. Initially, the package was oging 30 m/s. If it was to suddenly drop to zero, then the helicopters speed would have to instanteneously increase and that doesn't make sense.
     
  13. Jan 19, 2005 #12
    Well, what's strange is that the displacement of the crate is in the upwards direction.
     
  14. Jan 19, 2005 #13
    Not at all. Initially it has a velocity of 30 m/s UPWARDS. The acceleration is -9.8 m/s^2. In 2 seconds, the velocity will decrease by 19.6 m/s. Which means after the 2 seconds, its velocity is 10.4 m/s, but it is still UPWARDS. The velocity was always upwards, so the package itself must have travelled upwards.
     
  15. Jan 19, 2005 #14
    Yeah, Mathematically it looks right. But my intuition says the displacement should be downwards.
     
  16. Jan 19, 2005 #15
    Yea, that may seem to be what happens. But infact, that isnt. In real life the package will go upwards for a little bit. However, you would probably just never notice this because the distance it travels upwards or the time it takes is usually very small.
     
  17. Jan 20, 2005 #16

    Galileo

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    Can a helicopter really rise 30 m/s? That's 108 km/h; the package would still be rising after 2 s.
     
  18. Jan 20, 2005 #17
    Exactly. That's why I got confused a little. :rofl:
     
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