Question

During takeoff, an airplane climbs with a speed of 150 m/s at an angle of 45 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

Answer 1

106.1 m/s

Step-by-step explanation:

The shadow of the plane is moving at the same velocity as the horizontal component of the airplane's velocity.

The horizontal component of the airplane's velocity is

`v_x = v cos \theta`

where

v = 150 m/s is the velocity of the airplane

`\theta=45^(\circ)` is the angle between the airplane's velocity and the horizontal

Substituting,

`v_x = (150 m/s) cos 45^(\circ) = 106.1 m/s`

So, the shadow is moving at 106.1 m/s as well.