An airplane flies due north at 260 km/h with respect to the air. There is a wind blowing at 60 km/h to the northeast with respect to the ground. What are the plane's speed and direction with respect to the ground?

To find the plane's speed and direction with respect to the ground, we need to combine the vector components of the plane's velocity and the wind velocity.

The plane's velocity with respect to the ground can be calculated by adding the velocity of the plane with respect to the air and the velocity of the wind with respect to the ground.

First, let's break down the velocities into their respective components:

- The plane's velocity with respect to the air is due north, which means it has a vertical component of 260 km/h and a horizontal component of 0 km/h.

- The wind's velocity with respect to the ground is blowing 60 km/h to the northeast. To find the horizontal and vertical components, we can split the velocity into its northward and eastward components. Using trigonometry, we can determine that the wind has a vertical component of 60 km/h * sin(45 degrees) ≈ 42.4 km/h and a horizontal component of 60 km/h * cos(45 degrees) ≈ 42.4 km/h.

Next, we add the corresponding components together:

- The vertical component of the plane's velocity (260 km/h) remains unchanged, as the wind is not affecting it in that direction.

- The horizontal component of the plane's velocity (0 km/h) is added to the horizontal component of the wind's velocity (42.4 km/h).

So, the resulting velocity with respect to the ground has a vertical component of 260 km/h and a horizontal component of 0 km/h + 42.4 km/h = 42.4 km/h.

Using the Pythagorean theorem, we can calculate the magnitude of this resulting velocity:

Magnitude = sqrt(Vertical component^2 + Horizontal component^2)
= sqrt(260 km/h^2 + 42.4 km/h^2)
≈ sqrt(67600 km^2/h^2 + 1793.76 km^2/h^2)
≈ sqrt(69493.76 km^2/h^2)
≈ 263.61 km/h

So, the plane's speed with respect to the ground is approximately 263.61 km/h.

To determine the direction, we can use trigonometry again. The angle θ that the velocity makes with the north can be calculated as:

θ = arctan(Vertical component / Horizontal component)
= arctan(260 km/h / 42.4 km/h)
≈ arctan(6.132)

Using a calculator, we find that θ ≈ 81.8 degrees.

Since the plane is flying due north with a positive velocity, the direction with respect to the ground is the angle south of due east. Therefore, the plane's direction with respect to the ground is approximately 90 degrees + 81.8 degrees ≈ 171.8 degrees (or approximately south-southeast).