A pilot must fly his plane due north to reach his destination. The plane can fly at 300 km/h in still air. A wind is blowing out of the northeast at 90 km/h. (a) What is the speed of the plane relative to the ground

229.5 km/h

Explanation:

Given that

Velocity of the plane relative to the ground = V.pg = ?

Velocity of air relative to the ground = V.ag = 90 km/h

Velocity of the plane relative to air = V.pa = 300 km/h

V.pg = V.ag + V.pa

Where

V.ag = 90 km/h

= -90 cos 45i - 90 sin 45j

V.pa = 300 km/h

= 300 cos θi + 300 sin θj

V.pg = [-90 cos 45i - 90 sin 45j] + [300 cos θi + 300 sin θj], on rearranging and collecting like terms

V.pg = [300 cos θ - 90 cos 45]i + [300 sin θ - 90 sin 45]j

If vector i = 0, then

300 cos θ - 90 cos 45 = 0

300 cos θ = 90 cos 45

cos θ = 90 cos 45 / 300

cos θ = 63.6396 / 300

cos θ = 0.2121

θ = cos^-1 0.2121

θ = 77.75

We then use this θ to find vector j

V.pg = 300 sin θ - 90 sin 45

V.pg = 300 sin77.75 - 90 sin45

V.pg = 293.1693 - 63.6396

V.pg = 229.5297 km/h

a

explanation:

because they are cartillidge

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