Question

A man of height $2$ metre walks at uniform speed of $3$ metre per second away from the lamp post of height $5$ metre. Then the rate at which the length of his shadow increases is $\text{m/sec}$

Answer 1

Let $AB$ be the height lamp post and $MN$ be the height of the man.
Its shadow will be $y\text{m}$.
We know that,
$AB=5\text{m},MN=2\text{m}$
Assuming,
$AM=x,MC=y$
Then $\frac{dx}{dt}=3\text{m/s}$

Since, $△ABC$ and $△MNC$ are similar.
$\therefore \frac{MN}{AB}=\frac{MC}{AC}$
$\Rightarrow \frac{2}{5}=\frac{y}{x+y}$
$\Rightarrow y=\frac{2}{3}x$
Differentiating w.r.t. to $t$, we get
$\frac{dy}{dt}=\frac{2}{3}\times \frac{dx}{dt}$ $\Rightarrow \frac{dy}{dt}=\frac{2}{3}\times 3=2\text{m/sec}$