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Dog Sledding

  1. Jan 25, 2005 #1
    A cargo plane is dropping supplies to a group of people dogsledding acoss Antarctica. The plance drops the package when it is directly above the group of people at an altitude of 1950m. The dogsled travels at 5.68m/s and reaches the supplies in 6.70mins. What was the speed of the cargo plane?

    well I`m assuming that the plane and dogsled are traveling in opposite directions. So lets make the dogsled travel in the (+)ve direction and the plane traveling in the (-)ve direction.

    Xo = o
    X = ??
    to = 0s
    t = 402s
    Vdogsled = 5.68m/s
    Vplane = ???

    and then I used this equation: X = Xo + Vot to determine X
    X = 0 + 5.68m/s(402s)
    X = 2283.36m

    then using tan^-1 (theta) I found the angle which is 40.50degrees

    but now I'm stuck not quite sure what to do next - not quite sure if I'm on the right tracl :uhh:
     
  2. jcsd
  3. Jan 25, 2005 #2
    I think you are on the right track. But why would you assume the plane and sleddge are travelling in different directions?

    From the point you are now, what if you calculated how long it takes for the package to dropp to the ground? I think you can assume free fall.
     
  4. Jan 25, 2005 #3
    alright - I'll try that
     
  5. Jan 25, 2005 #4

    HallsofIvy

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    If "the plane and dogsled are traveling in opposite directions." and the plane dropped the supplies when it was directly overhead, the dogsled would go away from the supplies!
     
  6. Jan 25, 2005 #5
    i guess that makes sense - see I thought of it as the plane would drop the supplies directly over the sled but the supplies dont drop straight down but dop at an angle so i drew a triangle. but i guess plane and sled going the same way makes more sense;)
     
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