A man 1.8 m tall is walking at a rate of 1.5 m/s away from a streetlight. It is found that the length of his shadow is increasing at a rate of 0.9 m/s. How high above the ground is the streetlight?

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To find the height of the streetlight, we can set up a proportion using similar triangles.

Let's define the height of the streetlight as "h" and the length of the man's shadow as "x".

Since we have a similar triangle formed by the man, his shadow, and the streetlight, we can write the following proportion:

(man's height) / (man's shadow length) = (streetlight's height) / (streetlight's shadow length)

or

1.8 / x = h / (x + 1.8)

Now, to find the rate of change of the streetlight's height, we differentiate both sides of the equation with respect to time:

d(1.8) / dt = [d(h) / dt * (x + 1.8) - h * d(x) / dt] / (x + 1.8)^2

Since the man is walking away from the streetlight, the rate at which his shadow is increasing (0.9 m/s) is equal to dx / dt. Also, the man's height (1.8 m) and the length of his shadow (x) are constants, so their rates of change are zero.

Substituting the given information into the equation:

0 = [dh / dt * (x + 1.8) - h * 0.9] / (x + 1.8)^2

Simplifying the equation:

0 = dh / dt - 0.9h / (x + 1.8)

Now we have an equation relating the rate of change of the streetlight's height (dh / dt), the height of the streetlight (h), and the length of the shadow (x). But we need more information to solve for the height of the streetlight.

If we know the length of the man's shadow (x), we can substitute it into the equation and solve for the rate of change of the streetlight's height (dh / dt). Since we don't have that information, we cannot find the exact height of the streetlight.

Therefore, we cannot determine the height of the streetlight without additional information.

If the man's distance is x, and the shadow's length is s, and the light is at height h, then using similar triangles,

1.8/s = h/(x+s)
or,
1.8x + 1.8s = hs
1.8x = s(h-1.8)

1.9 dx/dt = (h-1.8) ds/dt

Now plug in your numbers and solve for h.