A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 2.3 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

Explanation:

Given the data in the question;

Let x represent the man's distance from building,

initially x = 1m2

dx/d t= -2.3 m/s

Also Let y represent shadow height

so we determine dy/dt when x is 4m from the building

form the image description of the problem, we see two-like triangles with the same base and height ratios

so

 2 / (12-x) = y / 12

24 = y(12 - x )

y = 24 / (12-x)

dy/dt = 24/(12-x)² × dx/dt

Now at x = 4,

we substitute

dy/dt will be;

⇒ 24/(12 - 4)² × -2.3

= 24/64 - 2.3

= 0.375 × -2.3

dy/dt = - 0.8625 m/s

Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

pe = weightxheight = 2000x1000=2,000,000joules

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong