How Fast Does a Hawk's Shadow Move When Diving?

[garcia1]

Homework Statement

When the Sun is directly overhead, a hawk
dives toward the ground at a speed of
6.04 m/s.
If the direction of his motion is at an angle
of 74.1 below the horizontal, calculate the
speed of his shadow along the ground.
Answer in units of m/s.

Homework Equations

My equation I used was

X(hypotenuse) = 6.04m/s / sin(74.1)

The Attempt at a Solution

My answer was 6.28 m/s, by trying to solve this problem via treating the answer as a hypotenuse between the known downward speed and the horizontal

 

[gneill]

Hi garcia1, welcome to Physics Forums.

It seems that the hawk's speed is along the trajectory of his 74.1 degree incline. With the Sun directly overhead, its shadow should be following along directly below him on the ground. Perhaps you should draw a diagram of the situation (in profile) to see what's happening.

 

[garcia1]

I took your advice and drew this diagram, realizing that the steep angle of the hawk's trajectory would make its downward speed the vertical and the speed of it's shadow along the ground as the horizontal. I came up with an answer of 1.72 m/s, using the following equation:

tan(74.1) = opp/adj = 6.04m/s / X

By solving for X, I got this answer of 1.72 m/s. This seems reasonable, since the horizontal is the smallest side of the right triangle I can make from these variables, but my answer still appears wrong on my homework. Any further thoughts?

 

[gneill]

The hawk's speed is along the trajectory, not vertical. If you draw the triangle, its speed is along the hypotenuse.

 

[rwisz]

direction.


I would like to provide a more thorough explanation for the solution to this problem. To calculate the speed of the hawk's shadow, we need to use vector components. The hawk's motion can be broken down into two components: the horizontal component and the vertical component. The horizontal component is equal to the speed of the hawk, which is 6.04 m/s. The vertical component is equal to the speed of the hawk multiplied by the sine of the angle, which is 6.04 m/s * sin(74.1) = 5.80 m/s.

The speed of the shadow is determined by the horizontal component, as the shadow will move parallel to the ground. Therefore, the speed of the shadow is also 6.04 m/s. This is because the horizontal component of the hawk's motion is the same as the horizontal component of the shadow's motion.

In summary, the speed of the hawk's shadow along the ground is also 6.04 m/s, which is equal to the horizontal component of the hawk's motion. It is important to consider both the horizontal and vertical components of motion when solving vector problems.