A balloon leaves the earth at a point  $A$ and rises vertically at uniform speed.  At the end of 2 minutes, John finds the angular elevation of the balloon as  $60°$.  If the point at which John is standing is  $150m$ away from point  $A$, what is the speed of the balloon?

Let $C$ be the position of John

Let  $A$ be the position at which balloon leaves the earth and

$B$ be the position of the balloon after 2 minutes

Given that $CA=150m$

$\angle BCA=60°$

$\tan 60°=\frac{BA}{CA}$

$\Rightarrow BA=CA\tan 60°$

$\Rightarrow BA=150\sqrt{3}$

i.e., the distance travelled by the balloon  $=150\sqrt{3}m$

time taken $=2\min =2\times 60=120\sec .$

Speed  $=\frac{\text{Distance}}{\text{Time}}$

$=\frac{150\sqrt{3}}{120}$

$=\frac{5\sqrt{3}}{4}$ m/sec.

$=\frac{5\times 1.73}{4}$

Speed  $=2.16m/\sec .$