Question

A pilot wants to fly on a bearing of 67.3°. By flying due east, he finds that a 51-mph wind, blowing from the south, puts him on course. Find the airspeed of the plane. Find the ground speed of the plane.

Answer 1

Then, using vector addition, we have the plane's velocity, in the form {s, 0}, where s is the airspeed. This is because east is in the x direction. The wind has the form {0, windspeed}, since North is in the y direction. Adding, we get {s, windspeed}

Since we know windspeed is 51 mph, and we know that the added vector (which is actually the ground velocity vector), we can see for which airspeed s we get an angle of 67.3 degrees using tangent:


Then tan(90-67.3) = 51/s

s = 51/ tan(90-67.3) = 121.919 mph.

For the groundspeed, we take the magnitude of the groundspeed velocity vector.
|{s, 51}| = sqrt(121.919^2+51^2) = 132.231