A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A , with…

Let speed of boat is $u \mathrm{m}/\mathrm{s}$ and height of tower
is $h$ meter & distance $AB=x$ metre

$\therefore \mathrm{x}=\mathrm{hcot}30°-\mathrm{hcot}45°$

$\Rightarrow x=h(\sqrt{3}-1)$

$\therefore \mathrm{u}=\frac{\mathrm{x}}{20}=\frac{\mathrm{h}(\sqrt{3}-1)}{20} \mathrm{m}/\mathrm{s}$

$\therefore$ Time taken to travel from $B$ to $C$ (Distance
$=h$ meter)

$=\frac{\mathrm{h}}{\mathrm{u}}=\frac{\mathrm{h}}{\mathrm{h}\frac{(\sqrt{3}-1)}{20}}=\frac{20}{\sqrt{3}-1}=10(\sqrt{3}+1)\sec .$