Asked by Mike
                a pilot wishes to fly on a course bearing 45 and with a ground speed of 600 km per hour. If a wind is blowing from the southwest (bearing 315) at 80 km per hour. What must be the heading and air speed of the aircraft?
            
            
        Answers
                    Answered by
            Steve
            
    since the wind and the desired course are in the same direction, the plane just needs to have an air speed of 520 km/hr at 45
    
                    Answered by
            Henry
            
    Vp + 80km/h[315] = 600[45o].
Vp + 80*sin315 + i80*Cos315 = 600*sin45 + i600*Cos45,
Vp - 56.6 + i56.6 = 424.3 + i424.3,
Vp = 480.9 + i367.7 = 605.4km/h[52.6o] = Air speed and heading of the plane.
    
Vp + 80*sin315 + i80*Cos315 = 600*sin45 + i600*Cos45,
Vp - 56.6 + i56.6 = 424.3 + i424.3,
Vp = 480.9 + i367.7 = 605.4km/h[52.6o] = Air speed and heading of the plane.
                    Answered by
            Henry
            
    Vp + 80[45o] = 600[45o].
Vp + 80*sin45 +i80*Cos45 = 600*sin45 + i600*Cos45,
Vp + 56.6 + i56.6 = 424.3 + i424.3,
Vp = 367.7 + i367.7 = 520 km/h[45o]. = Air speed and heading of the plane.
Note: If the wind is blowing from southwest, the bearing is 225o Not 315o.
  
    
Vp + 80*sin45 +i80*Cos45 = 600*sin45 + i600*Cos45,
Vp + 56.6 + i56.6 = 424.3 + i424.3,
Vp = 367.7 + i367.7 = 520 km/h[45o]. = Air speed and heading of the plane.
Note: If the wind is blowing from southwest, the bearing is 225o Not 315o.
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