During takeoff, an airplane climbs with a speed of 180m/s at an angle of 42 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. The sun is shining 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?)

Question

During takeoff, an airplane climbs with a speed of 180m/s at an angle of 42 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. The sun is shining 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?)

Answer 1

v upwards=180*sin42

v horizon=180*cos42

Answer 2

To find the magnitude of the horizontal component of the plane's velocity, we need to determine the horizontal speed of the plane.

Given that the plane's velocity can be represented by a vector with a speed of 180 m/s and an angle of 42 degrees above the horizontal, we can use trigonometry to find the horizontal component.

The horizontal component can be determined using the cosine function, which relates the angle to the adjacent side of a right triangle. In this case, the horizontal component is the adjacent side, and the hypotenuse is the plane's velocity.

So, to find the horizontal component, we use the formula:

Horizontal Component = Velocity * cos(Angle)

Substituting the given values:

Horizontal Component = 180 m/s * cos(42 degrees)

Using a scientific calculator, we can calculate the cosine of 42 degrees, which is approximately 0.7431.

Therefore:

Horizontal Component = 180 m/s * 0.7431

Horizontal Component = 133.558 m/s

So, the magnitude of the horizontal component of the plane's velocity is approximately 133.558 m/s.