Question

The height of a hill is $3000$ metres. From a point $P$ on the ground the angle of elevation of the top of the hill is ${60}^o$. A balloon is moving with constant speed vertically upwards from $P$. After $5$ minutes of its movement, a person sitting in it observes the angle of elevation of the top of the hill as ${30}^o$. What is the speed of the balloon?

• A. $62km/hr$
• B. $13km/hr$
• C. $24km/hr$
• D. $43km/hr$

Answer 1

Answer: (C)

$24km/hr$

height of hill $=3000$ m
height of balloon $=h$
Angle of elevation of hill from ground = ${60}^{\circ }$
Angle of elevation of hill from balloon = ${30}^{\circ }$
Let the horizontal distance be $d$
Now, When the person is on ground,
$\tan 60=\frac{3000}{d}$
$d=\frac{3000}{\sqrt{3}}=1000\sqrt{3}$
When the person is in the balloon,
$\tan 30=\frac{3000-h}{1000\sqrt{3}}$
$\frac{1}{\sqrt{3}}=\frac{3000-h}{1000\sqrt{3}}$
$1000=3000-h$
$h=2000$ m $=2$ km
$Speed=\frac{distance}{time}$
$Speed=\frac{2}{\frac{5}{60}}$
$\therefore Speed=24km/hr$