Question

A man $1.6m$ high walks at the rate of $30m/\min$ away from a lamp which is $4m$ above ground. How fast is the man's shadow lengthening?

• A. 22 m / min
• B. 20 m / min
• C. 15 m / min
• D. 25 m / min

Answer 1

Answer: (B)

Let $PQ=4m$ be the height of pole and, $AB=1.6m$ be height of man.
Let the end of shadow and it is at a distance of $l$ from $A$ when the man is at a distance $x$ from $PQ$ at some instant.
Since, $\Delta PQR$ and $\Delta ABR$ are similar,
we have, $\frac{PQ}{AB}=\frac{PR}{AR}$
$\Rightarrow \frac{4}{1.6}=\frac{x+l}{l}$
$\Rightarrow 2x=3l$
$\Rightarrow \frac{2dx}{dt}=3\cdot \frac{dl}{td}\{$ given $\frac{dx}{dt}=30m/\min \}$
$\Rightarrow \frac{dl}{dt}=\frac{2}{3}\times 30m/\min =20m/\min$