Question

SCALCET8 3.9.015. A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along a straight path. How fast is the tip of his shadow moving when he is 35 ft from the pole

Answer 1

`X=6.67ft/s`

Explanation:

From the question we are told that:

Height of pole
`H_p=15`

Height of man
`h_m=6ft`

Speed of Man
`\triangle a =4ft/s`

Distance from pole
`d=35ft`

Let

Distance from pole to man=a

Distance from man to shadow =b

Therefore

`(a+b)/(15)=(b)/(6)`

`6a+6b=15y`

`2a=3b`

Generally the equation for change in velocity is mathematically given by

`2(\triangle a)=3(\triangle b )`

`2*4=3(\triangle b)`

`\triangle a=(8)/(3)`

Since

The speed of the shadow is given as

`X=\triangle b+\triangle a`

`X=4+8/3`

`X=6.67ft/s`