Question

A man $1.5m$ tall walks away from a lamp post $4.5m$ high at a rate of $4km/hr$(i)How fast is his shadow lengthening?(ii)How fast is the farther end of shadow moving on the pavement? ?

Answer 1

By similarity of triangle

$\frac{4.5}{y}=\frac{1.5}{y-x}$

$4.5y-1.5y=4.5x$

$3y=4.5x$

$y=\frac{4.5x}{3}$

$\frac{dy}{dt}=\frac{4.5}{3}\frac{dy}{dt}$

$\frac{dy}{dt}=\frac{4.5}{3}\ast 4000=6000m$