Question

a man 150 cm tall walks away from a source of light situated at the top of a pole 5 m high at the rate of .7 m/s .find the rate at which the tip of his shadow is moving when he is 2 m away from the pole

Answer 1

Answer:

The rate of which shadow is moving is 2 m/s .

Step-by-step explanation:

Given as :

The height of man = CD = 150 cm = 1.5 m

The height of pole = AB = 5 m

The rate of man = 0.7 m/s

Horizontal distance between pole and man = AC = x m

Total distance of pole = AO = y m

According to question

from similar Triangle  AOB and COD

$\dfrac{AB}{CD}$  = $\dfrac{AC}{CO}$

Or, $\dfrac{5 m}{1.5 m}$ = $\dfrac{y}{y-x}$

Or, $\dfrac{10}{3}$ = $\dfrac{y}{y-x}$

or, 10 y - 10 x = 3 y

Or, 7 y = 10 x

or, $\dfrac{x}{y}$ = 0.7

∴   x = $\dfrac{0.7}{0.7}$

i.e  x = 1 m/s

So, The rate at which shadow is moving for 2 m = 1 m/s × 2 = 2 m/s

Hence, The rate of which shadow is moving is 2 m/s . Answer