A street light is mounted at the top of a 19-ft-tall pole. A man 5.5 feet tall walks away from the pole with a speed of 14 ft/s along a straight path. How fast is the tip of his shadow moving when he is 100 feet from the pole?

If the man is x feet from the pole, and his shadow has length s, then by similar triangles,

(x+s)/19 = s/5.5
19s = 5.5(x+s)
13.5s = 5.5x
s = .407x
The tip of the shadow is at distance x+s = 1.407x

so, it is moving with speed

d(x+s)/dt = 1.407 dx/dt
= 1.407*14 = 19.7 ft/s

To solve this problem, we need to use similar triangles and the concept of related rates.

Let's label the variables:
h = height of the man
x = distance from the man to the base of the pole
y = length of the shadow
θ = angle of elevation from the tip of the shadow to the top of the pole

We know that the height of the pole is 19 feet and the height of the man is 5.5 feet. The distance from the man to the base of the pole is given by x, and the length of the shadow is given by y.

Based on the concept of similar triangles, we can write the following ratios:

h / x = (h + y) / y

Substituting the given values:
5.5 / x = (5.5 + y) / y

Simplifying this equation, we get:
5.5y = (5.5 + y)x

To find the rate of change of the tip of the shadow with respect to time, we differentiate both sides of the equation with respect to time:

5.5(dy/dt) = (dx/dt)(5.5 + y) + x(dy/dt)

Now we can substitute the given values:
5.5(dy/dt) = (dx/dt)(5.5 + y) + x(dy/dt)
5.5(dy/dt) = (14)(5.5 + 100) + 100(dy/dt)
5.5(dy/dt) = 777 + 100(dy/dt)

Rearranging the equation, we get:
4.5(dy/dt) = 777

Therefore, dy/dt = 777 / 4.5
dy/dt ≈ 173 ft/s

So, the tip of the man's shadow is moving at a speed of approximately 173 ft/s when he is 100 feet from the pole.