During takeoff, an airplane climbs with a speed of 140 m/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 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?)
horizontal component of the plane's velocity = 140 cos42 m/s
To determine how fast the shadow of the plane is moving along the ground, we need to find the magnitude of the horizontal component of the plane's velocity.
The velocity vector of the plane is given by the speed (140 m/s) and the angle of 42 degrees above the horizontal.
To find the horizontal component, we can use trigonometry. The horizontal component is given by the equation:
Horizontal component = Velocity * cosine(angle)
Substituting the given values:
Horizontal component = 140 m/s * cosine(42 degrees)
To find the magnitude of the horizontal component, we evaluate the right side of the equation:
Horizontal component = 140 m/s * 0.7431
Simplifying:
Horizontal component = 103.834 m/s
Therefore, the magnitude of the horizontal component of the plane's velocity (the speed of the shadow on the ground) is approximately 103.834 m/s.