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[SOLVED] The Angle of the Launch of a Rocket against a Strong Wind
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?
Total Time- t=((2(initial velocity)sin(j))/g)
Position Function- x=initial position+(initial velocity)t
f=ma
weight=(mass)(gravity)
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
x=Vcos(j)((2Vsin(j))/g)+.5(-8650/1210)((2Vsin(j))/g)^2
x=(2(V^2)cos(j)sin(j))/9.81+(-8650/2420)(4V^2(sin(j))^2)/g^2
x=(V^2/g)sin(2j)-25.256411(sin(j))^2
x=736.4934sin(2j)-25.2564(sin(j))^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.
Homework Statement
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?
Homework Equations
Total Time- t=((2(initial velocity)sin(j))/g)
Position Function- x=initial position+(initial velocity)t
f=ma
weight=(mass)(gravity)
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
x=Vcos(j)((2Vsin(j))/g)+.5(-8650/1210)((2Vsin(j))/g)^2
x=(2(V^2)cos(j)sin(j))/9.81+(-8650/2420)(4V^2(sin(j))^2)/g^2
x=(V^2/g)sin(2j)-25.256411(sin(j))^2
x=736.4934sin(2j)-25.2564(sin(j))^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.