A man of height 2m walks directly away from a lamp of h

SOLUTION

Let be the lamp and PQ be the man and OQ=x metre be his shadow and let MQ =y metre.

$\therefore \frac{dy}{dt}$=speed of the man
=3 m/s (given)
$\because \Delta OPQ$ and $\Delta OLM$ are similar
$\therefore \frac{OM}{OQ}=\frac{LM}{PQ}\Rightarrow \frac{x+y}{x}=\frac{5}{2}\Rightarrow y=\frac{3}{2}x\therefore \frac{dy}{dt}=\frac{3}{2}\frac{dx}{dt}\Rightarrow 3=\frac{3}{2}\frac{dx}{dt}\Rightarrow \frac{dx}{dt}=2m/s$