An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 10.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination? a. 4.78 degree E of N
b. 4.76 degree E of N
c. 85.2 degree W of N
d. 4.78 degree W of N
e. 4.76 degree W of N

A

Explanation:

Given that

V' = speed of flight of the plane, 120 m/s

V(0) = speed of the wind, 10 m/s

From a rough sketch, we can see that the flight takes place in shape of a right angled triangle of which both the speed of the plane and that of the wind are adjacent to each other. Thus, invoking Pythagoras' rule, we have

Cos Φ = V(0)/V', making Φ the subject of formula, we have

Φ = Cos^-1 [V(0)/V']

On substituting the values, we have

Φ = Cos^-1 [10/120]

Φ = Cos^-1 0.083

Φ = 85.2 to 1 decimal place

The final piece of the jigsaw is subtracting this angle from 90, so we have

90 - 85.2 = 4.78°

Since the wind is blowing at a direction due West during the flight, then the plane would have to fly due East instead.

like poles repel so keep pushing north to north and repels funny to see this

k2s for potassium sulphide