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

Answer:

$\dfrac{dL}{dt}=5.82 \ ft/s$

Explanation:

given,

street light height = 13 ft

man height = 6.3 ft

speed of the man = 3.5 ft/sec

$\dfrac{H}{L} = \dfrac{h}{l}$

$\dfrac{H}{L} = \dfrac{h}{L-x}$

$\dfrac{L}{H} = \dfrac{L-x}{h}$

hL = H(L-x)

hL = HL-Hx

$L = \dfrac{Hx}{H-h}$

$L = \dfrac{13x}{13-6.3}$

L = 1.94 x

$\dfrac{dL}{dt}=\dfrac{dL}{dx}\dfrac{dx}{dt}$

$\dfrac{dL}{dt}=1.94\times 3$

$\dfrac{dL}{dt}=5.82 \ ft/s$