The plane is trying to fly due east, but is lifted upward by the wind.
This creates a triangle with a hypotenuse of 490 and an opposite side of 80.
Let the angle of elevation be theta.
sin = opposite/hypotenuse = 80/490
theta = arcsin (80/490) ~ 9.4 degrees
Ground speed = (490) cos(9.4) ~ 483.4 km/hr
Kathy P.
tutor
OK. If you turn this problem into 3D with the x-axis for West-East, the y-axis for (North-South), and the z-axis for elevation, then...
The plane is represented by a rectangle in the xz-plane with diagonal 490 and z-axis = 80. Angle of elevation is Theta = ArcSin(80/490) = 9.4 degrees. x-compnent is 490*cos(9.4) = 483.4. The 3D vector for the plane is <483.426,0, 80>.
The wind is blowing NW so it is represented as a square with diagonal 110. x = y = 110*Cos(45) = 77.78
The 3D Vector for the wind is <-77.78, -77.78, 0>
The total is the Plane Vector + Wind Vector
Total = <483.43, 0, 80> + <-77.78, -77.78, 0> = <405.6, -77.78, 80>
From the ground, the 2D vector is <405.6, -77.78>
Upward movement would not be detected.
Speed from the ground = Sqrt( 406^2 + 77.8^2) = 412.4 km/hr
Angle = ArcTan(77.8/406) = 10.8 degrees
Report
12/26/21
William J.
The answer is making the incorrect assumption that the rate of climb is caused by the wind. The wind is the third factor in the problem. Factor (1.) Rate of Climb, Factor (2.) Air Speed, Factor (3.) Wind Speed and Wind Direction. The answer requires further work to address the third factor. The problem clearly states three factors and only the first two have been addressed.12/26/21