Airplane flying at 400 mph at an angle of 30 deg encounters a wind. The resultant velocity of the airplane is 475.3 mph at an angle of 27.18 deg. What was direction of the wind.
I set this up as tan (27.18) = (Ry/Rx)
But the total Ry contains the y vector of the plane and the y vector of the wind, and since I don't know either wind magnitude nor direction I end up with 2 unknowns in both Ry and Rx.
Then I thought about using law of cosines, but again, come up with one equation, 2 unknowns.
Any thoughts?
Best Answer
It may be easier to treat the direction of the plane as along one axis, like the $x$ axis:
$$\vec{v_0} = 400 \hat{x}.$$
Then the new velocity is
$$\vec{v} = 475.3 (\cos (-2.82^{\circ}) \hat{x} + \sin (-2.82^{\circ}) \hat{y}) \approx 474.72 \hat{x} - 23.38 \hat{y}.$$
So the speed of the wind is
$$\sqrt{74.72^2 + (-23.38)^2} \approx 78.29 \text{ MPH}$$
and its direction is
$$\arctan(-23.38/74.72) + 30^{\circ} \approx -17.37^{\circ} + 30^{\circ} = 12.63^{\circ}.$$