A plane points its nose due north and flies at a speed of 71 m/s relative to the air. The plane enters a strong wind that blows at a speed of 23 m/s due west relative to the ground. What is the angle of the plane's trajectory according to an observer standing on the ground? Assume that we measure the angle in degrees with respect to due east, so 0° would be due east, 90° would be due north, etc.

Question

A plane points its nose due north and flies at a speed of 71 m/s relative to the air. The plane enters a strong wind that blows at a speed of 23 m/s due west relative to the ground. What is the angle of the plane's trajectory according to an observer standing on the ground? Assume that we measure the angle in degrees with respect to due east, so 0° would be due east, 90° would be due north, etc.

Answer 1

Vp = Vo + Vw = 71i - 23 = -23 + 71i.

Tan Ar = 71/-23 = -3.08696
Ar = -72.0o S. of E.
A = -72.0 + 180 = 108o CCW.

Answer 2

To determine the angle of the plane's trajectory, we can use vector addition.

The plane's velocity relative to the ground can be found by adding its velocity relative to the air with the velocity of the wind.

The plane's velocity relative to the air is 71 m/s due north.

The wind's velocity relative to the ground is 23 m/s due west.

To find the plane's velocity relative to the ground, we need to add the magnitudes and directions of these velocities.

Since the plane is flying north (0°), and the wind is blowing west (270°), we can represent these velocities in terms of their components along the x and y axes.

The component of the plane's velocity in the x-direction (east-west) is 0 m/s, as it is pointing due north.

The component of the plane's velocity in the y-direction (north-south) is 71 m/s.

The wind's component in the x-direction is -23 m/s (negative because it's blowing west) and in the y-direction is 0 m/s (no wind in the north-south direction).

Adding the x-components and the y-components separately, we get:

x-component = 0 m/s + (-23 m/s) = -23 m/s
y-component = 71 m/s + 0 m/s = 71 m/s

Now, we can use trigonometry to find the angle.

The angle of the plane's trajectory with respect to due east (0°) is the inverse tangent (arctan) of the y-component divided by the x-component:

angle = arctan(y-component / x-component) = arctan(71 m/s / -23 m/s)

Using a calculator to find the arctan of (71 / -23), we get:

angle ≈ -71.3°

Since angles are measured counterclockwise from due east, a negative sign indicates a clockwise angle rotation. Therefore, the angle of the plane's trajectory with respect to due east is approximately 71.3° clockwise.