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Relative Velocity of a Helicopter

  1. Oct 13, 2008 #1
    1. The problem statement, all variables and given/known data
    Average Wind Velocity 38km/h [25 degrees N of E]
    Helicopter needs to achieve 91km/h [17 degrees W of N]


    2. Relevant equations
    r=sqrt(x^2+y^2)


    3. The attempt at a solution

    y=38sin25 + 91sin98 = 106.17
    x=38cos25 - 91cos98 = 47.1

    r= 116.15

    answer should be 94km/h
     
  2. jcsd
  3. Oct 14, 2008 #2

    alphysicist

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    Homework Helper

    Hi vsharma88,

    You have a bit of an error with the trig functions, but the most important thing here is how you have written down your equations. It appears that you are adding the 38km/h wind velocity to the 91km/h velocity to find a new resultant.

    However, the way I read the problem indicates that the 91km/h is the resultant. So the question here is what do you have to add to the wind velocity so that the helicopter goes 91km/h in the specified direction?
     
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