a gymnast tumbles forward 4.0m does cartwheels to the left for 6.0 m and climb a vertical rope to a height of 3.0m. What is the magnitude of the plane's displacement from the airport?

To find the magnitude of the plane's displacement from the airport, we can use the concept of vector addition.

First, let's break down the gymnast's movements into vector components. We'll use a coordinate system where the positive x-axis is forward, the positive y-axis is left, and the positive z-axis is upward.

The gymnast tumbles forward for 4.0 m, which corresponds to a displacement vector of (4.0, 0, 0) m.

Next, the gymnast does cartwheels to the left for 6.0 m, which corresponds to a displacement vector of (0, -6.0, 0) m.

Finally, the gymnast climbs a vertical rope to a height of 3.0 m, which corresponds to a displacement vector of (0, 0, 3.0) m.

To find the overall displacement, we can simply add these vectors together. Adding the x-components, we get 4.0 m. Adding the y-components, we get -6.0 m. Adding the z-components, we get 3.0 m.

Therefore, the overall displacement vector is (4.0, -6.0, 3.0) m.

To find the magnitude of this displacement vector, we can use the Pythagorean theorem in three dimensions:

Magnitude = sqrt(x^2 + y^2 + z^2)

Plugging in the values from the displacement vector, we have:

Magnitude = sqrt((4.0)^2 + (-6.0)^2 + (3.0)^2)

Simplifying and calculating, we find:

Magnitude ≈ 8.77 m

So, the magnitude of the plane's displacement from the airport is approximately 8.77 m.