During takeoff, an airplane climbs with a speed of 180 m/s at an angle of 34° 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?)
look at the triangle. It is clear that the horizontal component of the speed is 180 cos34°
To find the magnitude of the horizontal component of the plane's velocity, we first need to find the horizontal component of the velocity vector.
The velocity vector can be split into two components: horizontal and vertical. The horizontal component represents the motion in the horizontal direction, while the vertical component represents the motion in the vertical direction.
Given that the speed of the airplane is 180 m/s and the angle of climb is 34°, we can determine the horizontal component using trigonometry.
The horizontal component can be found by using the cosine of the angle:
Horizontal component = speed * cosine(angle)
Horizontal component = 180 m/s * cosine(34°)
To calculate this using a scientific calculator:
- Set the calculator to degree mode.
- Enter the value for cosine (34°).
- Multiply the result by the speed (180 m/s).
Using these steps, the calculation becomes:
Horizontal component = 180 m/s * cos(34°)
= 180 m/s * 0.829
= 149.22 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 149.22 m/s.