A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is 3h. The rate at which of the length of the man's shadow is increasing when he is at a distance 10 h from the base of the street lamp is:

Question

A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is 3h. The rate at which of the length of the man's shadow is increasing when he is at a distance 10 h from the base of the street lamp is: (A) v / 2 (B) v / 3 (C) 2 v (D) v / 6

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0c8e7ef1714fe75b37898319c4e9a0da-.png" /> (3 h-h/y)=(h/x) ⇒ (2 h/y)=(h/x) x=(y/2) ⇒ (d x/d t)=(1/2) (d y/d t) v text shodow =( v /2)

Q. A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is $3 h$ . The rate at which of the length of the man's shadow is increasing when he is at a distance $10 h$ from the base of the street lamp is:

$v /2$

$v /3$

$2 v$

$v /6$

$\frac{3 h - h}{y} = \frac{h}{x}$
$\Rightarrow \frac{2 h}{y} = \frac{h}{x}$
$x = \frac{y}{2}$
$\Rightarrow \frac{d x}{d t} = \frac{1}{2} \frac{d y}{d t}$
$v_{shodow} = \frac{v}{2}$

Q. A man of height 'h' is walking away from a street lamp with a constant speed 'v'. The height of the street lamp is 3h. The rate at which of the length of the man's shadow is increasing when he is at a distance 10h from the base of the street lamp is:

v/2

v/3

2v

v/6