You are traveling on an airplane. The velocity of the plane with respect to the air is 110.0 m/s due east. The velocity of the air with respect to the ground is 39.0 m/s at an angle of 30° west of due north.

1) What is the speed of the plane with respect to the ground?
2) What is the heading of the plane with respect to the ground? (Let 0° represent due north, 90° represents due east).
3)How far east will the plane travel in 1 hour?

1. Vr = 110m/s[0o] + 39[120o].

X = 110 + 39*Cos120 = 90.5 m/s.
Y = 39*sin120 = 33.8 m/s.
Vr = 90.5 + 33.8i = 96.6m/s[20.5o]N. of E.

2. A + 20.5 = 90.
A = 69.5o E. of N. = 69.5o CW from +Y-axis.

3. d = Vr*T*sin A = 96.6*3600*sin69.5 = 325,737 m. = 325.74 km.

I'm just gonna guess:

1) 100.038888m/s
2) 77.0000 degrees East of due North
3) 360140.040000m

To find the speed of the plane with respect to the ground, we can use vector addition.

1) The speed of the plane with respect to the ground is the magnitude of the resultant vector formed by adding the velocity of the plane with respect to the air and the velocity of the air with respect to the ground.

Given:
Velocity of the plane with respect to the air (Vpa) = 110.0 m/s due east
Velocity of the air with respect to the ground (Vag) = 39.0 m/s at an angle of 30° west of due north

To find the speed, we need to add the vectors:
Vpg = Vpa + Vag

To add the vectors, we can break each vector into its horizontal and vertical components.

The horizontal component of Vpa is 110.0 m/s, and the horizontal component of Vag can be found by multiplying Vag by the cosine of the angle:
Vag_horizontal = 39.0 m/s * cos(30°)

The vertical component of Vag can be found by multiplying Vag by the sine of the angle:
Vag_vertical = 39.0 m/s * sin(30°)

Now we can add the horizontal and vertical components of the vectors:

Vpg_horizontal = Vpa_horizontal + Vag_horizontal
Vpg_vertical = Vag_vertical

Using the Pythagorean theorem, we can find the magnitude of the resultant vector:

Vpg = sqrt(Vpg_horizontal^2 + Vpg_vertical^2)

Now you can substitute the values and calculate the speed of the plane with respect to the ground.

2) To find the heading of the plane with respect to the ground, we need to find the angle between the resultant vector (Vpg) and the north direction.

The heading can be found using the inverse tangent function:
Heading = arctan(Vpg_vertical / Vpg_horizontal)

Substitute the values into the equation to find the heading.

3) To find how far east the plane will travel in 1 hour, we can use the speed of the plane with respect to the ground from part 1 and multiply it by the duration of travel.

Distance = Speed * Time
Distance east = Speed * Time east

Substitute the values into the equation to find the distance traveled east.

By following these steps, you can find the answers to the given questions.