Question

A street light is at the top of a 13.0 ft. tall pole. A man 6.3 ft tall walks away from the pole with a speed of 3.5 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 48 feet from the pole?

Answer 1

`(dL)/(dt)=5.82 \ ft/s`

Step-by-step explanation:

given,

street light height = 13 ft

man height = 6.3 ft

speed of the man = 3.5 ft/sec

`(H)/(L) = (h)/(l)`

`(H)/(L) = (h)/(L-x)`

`(L)/(H) = (L-x)/(h)`

hL = H(L-x)

hL = HL-Hx

`L = (Hx)/(H-h)`

`L = (13x)/(13-6.3)`

L = 1.94 x

`(dL)/(dt)=(dL)/(dx)(dx)/(dt)`

`(dL)/(dt)=1.94* 3`

`(dL)/(dt)=5.82 \ ft/s`