A pilot flying from Honolulu to California flies at an airspeed of 300.0 m/s (\"

Question

A pilot flying from Honolulu to California flies at an airspeed of 300.0 m/s ("airspeed" - airplane's speed through the surrounding air) with her airplane pointed 67.0 degrees to the east of north. However, the air itself is moving with a 75.0-m/s windspeed blowing due east. When the pilot adds together the two vectors of her airspeed and the windspeed, she gets her groundspeed: the airplane's actual velocity relative to the ground below. (Even though these are called "speeds" by pilots, they are all vector velocities.) Using either the component method of the "triangle" method, find the magnitude and compass bearing (expressed in degrees to the east of north) of the airplane's groundspeed.

Explanation / Answer

East component of aircraft airspeed = (sin 67) x 300, = 276.15 m/sec.

North component = (cos 67) x 300, = 117.22m/sec.

Add east wind to east component, = 276.15 + 75, = 351.15m/sec. east.
Sqrt. (351.15^2 + 117.22^2) = 370.2m/sec. ground speed.

Direction = arctn. (351.15/117.22) = 71.54 deg., E of N.

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