During takeoff, an airplane climbs with a speed of 300 m/s at an angle of 36 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?)
consider what would happen if the plane were flying horizontally. The shadow would be moving exactly as fast as the plane.
What if the plane were flying vertically? Shadow would not move.
So, it's easy to see that if the plane is rising at an angle θ, the horizontal speed is 300 cosθ
To find the speed of the shadow of the plane moving along the ground, we need to determine the magnitude of the horizontal component of the plane's velocity.
The horizontal component of the plane's velocity can be found by finding the cosine component of the velocity vector.
Given:
Velocity of the plane (v) = 300 m/s
Angle (θ) = 36 degrees
To find the horizontal component (v_x) of the velocity vector:
v_x = v * cos(θ)
Substituting the given values:
v_x = 300 m/s * cos(36 degrees)
Using a calculator, we can find:
v_x ≈ 300 m/s * 0.809 = 243 m/s
Therefore, the shadow of the plane is moving along the ground with a speed of approximately 243 m/s.
To find the magnitude of the horizontal component of the plane's velocity, we need to decompose the velocity vector into its horizontal and vertical components.
Given:
Velocity of the airplane (v) = 300 m/s
Angle made by the velocity vector with the horizontal (θ) = 36 degrees
To find the horizontal component of the velocity (v_x):
v_x = v * cos(θ)
Using the given values, we can calculate:
v_x = 300 m/s * cos(36 degrees)
Now let's calculate it:
v_x = 300 m/s * cos(36 degrees)
v_x ≈ 300 m/s * 0.809
v_x ≈ 242.7 m/s
Therefore, the magnitude of the horizontal component of the plane's velocity, which determines the speed of the shadow along the ground, is approximately 242.7 m/s.