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Math Help - Need help with plane type of Questions

  1. #1
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    Need help with plane type of Questions

    Hello, these two questions are from my Calc and Vectors book
    1)A small airplane takes off at an airspeed of 180km/hr, at an angle of inclination of 14degrees, toward the east. A 15km/hr wind is blowing from the southwest. Determine the resultant ground velocity

    2)An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnnitudes.
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  2. #2
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    Quote Originally Posted by supersaiyan View Post
    Hello, these two questions are from my Calc and Vectors book
    1)A small airplane takes off at an airspeed of 180km/hr, at an angle of inclination of 14degrees, toward the east. A 15km/hr wind is blowing from the southwest. Determine the resultant ground velocity
    air vector parallel to the ground ... 180\cos(14)

    east component ... v_x = 180\cos(14) + 15\cos(45)

    north component ... v_y = 15\sin(45)

    resultant ground speed = \sqrt{v_x^2 + v_y^2}

    direction relative to east ... \theta = \arctan\left(\frac{v_y}{v_x}\right)
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    Quote Originally Posted by skeeter View Post
    air vector parallel to the ground ... 180\cos(14)

    east component ... v_x = 180\cos(14) + 15\cos(45)

    north component ... v_y = 15\sin(45)

    resultant ground speed = \sqrt{v_x^2 + v_y^2}

    direction relative to east ... \theta = \arctan\left(\frac{v_y}{v_x}\right)
    Thanks, but why didn't we also use v_y = 15\sin(45) +  180\sin(14)
    Why did we only use 15\sin(45)
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    Quote Originally Posted by supersaiyan View Post
    Thanks, but why didn't we also use v_y = 15\sin(45) +  180\sin(14)
    Why did we only use 15\sin(45)
    there is no component of the air vector in the y-direction.
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