Question

A man 6.00 ft tall approaches a street light 15.0 ft above the ground at the rate of 4.00 ​ft/s. How fast is the end of the​ man's shadow moving when he is 14.0 ft from the base of the​ light?

Answer 1

`(dx)/(dt) = 10 ft/s`

Step-by-step explanation:

As per given figure let say the tip of the shadow is at distance "x" from the base of the lamp

so here we have

`(x)/(15) = (x - y)/(6)`

so we have

`6x = 15 x - 15 y`

`15 y = 9 x`

now we have

`5(dy)/(dt) = 2(dx)/(dt)`

`5(4) = 2(dx)/(dt)`

`(dx)/(dt) = 10 ft/s`