During takeoff, an airplane climbs with a speed of 120 m/s at an angle of 33 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?)
Xo = Vo*Cos A = 120*Cos33 =
101 m/s
To find the magnitude of the horizontal component of the plane's velocity, we need to consider the given information. The speed of the plane is 120 m/s, and the angle above the horizontal is 33 degrees.
The velocity of the plane can be divided into two components: the horizontal component and the vertical component. The horizontal component is the part of the velocity that acts in the horizontal direction, while the vertical component is the part that acts in the vertical direction.
To find the magnitude of the horizontal component, we can use trigonometry. The horizontal component can be found by multiplying the speed of the plane by the cosine of the angle. In this case, the horizontal component can be calculated as:
Horizontal component = Speed × Cos(angle)
Using the given values:
Horizontal component = 120 m/s × Cos(33 degrees)
Now, we can calculate the magnitude of the horizontal component by evaluating the expression:
Horizontal component = 120 m/s × 0.8387
Therefore, the magnitude of the horizontal component of the plane's velocity, and hence the speed at which the shadow of the plane is moving along the ground, is approximately 100.64 m/s.