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?

Answer 1

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.

Answer 2

Thank you so so much!:)