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?
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)
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.posted by Steve
Thank you so so much!:)posted by Sisi