Question

A man of height 2m walks at uniform speed of 5km/hr away from a lamp post which is 6m high. Find the rate of which the length of his shadow increases.

Answer 1

Let $AB$ be the lamp post

and let $MN$ be the man of height $2$m

and let $AM=l$meter.

and $MS$ is the shadow of the man.

Let length of shadow $MS=s$

Given: Man walks at speed of $5$kmph

$\therefore \frac{dl}{dt}=5$kmph

We need to find rate at which the length of his shadow increases$=\frac{ds}{dt}$.

In $△ASB,\tan \theta =\frac{AB}{AS}=\frac{6}{l+s}$ ....$\left(1\right)$

In $△MSN,\tan \theta =\frac{MN}{MS}=\frac{2}{s}$ ....$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\frac{6}{l+s}=\frac{2}{s}$

$\Rightarrow 6s=2l+2s$

$\Rightarrow 4s=2l$

or $l=2s$

$\frac{dl}{dt}=2\frac{ds}{dt}$

$5=2\frac{ds}{dt}$ since given that $\frac{dl}{dt}=5$kmph

$\therefore \frac{ds}{dt}=\frac{5}{2}$kmph$=2.5$kmph