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Homework Help: The Angle of the Launch of a Rocket against a Strong Wind

  1. Feb 8, 2008 #1
    [SOLVED] The Angle of the Launch of a Rocket against a Strong Wind

    1. The problem statement, all variables and given/known data
    A 1210-kg rocket is launched with an initial velocity v = 85 m/s against a strong wind. The wind exerts a constant horizontal force F= 8650 N on the rocket. At what launch angle will the rocket achieve its maximum range in the up-wind direction?

    2. Relevant equations
    Total Time- t=((2(initial velocity)sin(j))/g)
    Position Function- x=initial position+(initial velocity)t

    3. The attempt at a solution

    Alright, first I broke the velocity into components. The velocity in the vertical direction is the initial velocity times sin(j). The velocity in the horizontal is the initial velocity times cos(j). The force of the wind is -8650 N and only effects the horizontal. I figured out the acceleration of the wind, -8650/1210 m/s^2. I then plugged all the information I had into the position function in order to maximize total distance.

    x=initial position+(initial velocity)t+.5(Fw/mg)t^2

    I then graphed this found took the max point. I found this to be 44.5088400. This answer was incorrect. Thus, I took the derivative with respect to j. As expected, I got the same result. I went and spoke with my TA's and they informed me that this was the correct method. Any help you could give me would be great.
  2. jcsd
  3. Feb 9, 2008 #2
    I did it by a different method and found the answer to be 27 degrees. Is my answer correct?
    Last edited: Feb 9, 2008
  4. Feb 9, 2008 #3
    I think you made a mistake. I got

    [tex] y=736.5sin2x-1073sin^2x[/tex]
  5. Feb 9, 2008 #4
    yeah, I got those numbers a few hours later. I'm not sure how exactly I managed, but I found that theta was equal to arctan (mg/Fwind)/2. The correct answer is close to 27 degrees.
  6. Feb 9, 2008 #5
    You will get that exact expression when you differentiate the expression in my previous post .
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