During takeoff, an airplane, climbs with a speed of 180 m/s at an angle of 34º above the horizontal. The sun is shining directly overhead. How fast is the shadow of the plane moving along the ground? (That is, what is the magnitude of the horizontal component of the plane's velocity?)

Xo = 180*cos34 = 149.2 m/s.

Xo = 180*cos34 = 149.2 m/s

To find the magnitude of the horizontal component of the plane's velocity, we need to determine the rate at which the shadow of the plane is moving along the ground.

Since the sun is shining directly overhead, the angle between the direction of sunlight and the horizontal is 90 degrees. Therefore, the angle between the direction of the shadow and the horizontal is also 90 degrees.

Now, we can use trigonometry to solve the problem. We know that the angle of climb above the horizontal is 34 degrees, so the angle between the plane's velocity vector and the horizontal is also 34 degrees.

The magnitude of the horizontal component of the plane's velocity can be found by using the cosine function:

cos(34°) = Adjacent / Hypotenuse.

In this case, the hypotenuse represents the plane's velocity of 180 m/s, and the adjacent side represents the horizontal component of the velocity.

So, the equation becomes:

cos(34°) = Adjacent / 180 m/s.

To find the value of the adjacent side, we rearrange the equation:

Adjacent = cos(34°) * 180 m/s.

Evaluating this expression:

Adjacent = cos(34°) * 180 m/s ≈ 149.55 m/s.

Therefore, the magnitude of the horizontal component of the plane's velocity, which represents the speed at which the shadow of the plane is moving along the ground, is approximately 149.55 m/s.