During takeoff, an airplane climbs with a speed of 110 m/s at an angle of 44 degrees above the horizontal. The speed and direction of the airplane constitute a vector quantity known as the velocity. 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?)
The plane's velocity vector is the hypotenuse of a right triangle. You know its magnitude and angle, and you need to find the base's magnitude.
To find the magnitude of the horizontal component of the plane's velocity, we need to use trigonometry.
Let's break down the information given:
- The airplane climbs with a speed of 110 m/s. This is the magnitude of the velocity vector.
- The angle the airplane makes with the horizontal is 44 degrees.
The velocity vector can be split into two components: the vertical component (representing the ascent of the airplane) and the horizontal component (representing the motion along the ground).
To find the magnitude of the horizontal component, we need to find the cosine of the angle between the velocity vector and the ground.
cos(θ) = adjacent / hypotenuse
In this case, the adjacent side is the horizontal component of the velocity, and the hypotenuse is the magnitude of the velocity (110 m/s).
cos(44°) = horizontal component / 110 m/s
Now, let's solve for the horizontal component:
horizontal component = 110 m/s * cos(44°)
Using a calculator, we can calculate the value of cos(44°) to be approximately 0.7193.
horizontal component ≈ 110 m/s * 0.7193
horizontal component ≈ 79.12 m/s
Therefore, the magnitude of the horizontal component of the plane's velocity, which represents the speed of the shadow along the ground, is approximately 79.12 m/s.