Question

The angle of elevation of a hot-air balloon climbing vertically changes from 30° at 10:00 AM to 60° at 10:02

AM as observed from a point 300 m away from the take-off point. What is the upward speed of the balloon?

A)2.18 m/s B)2 m/s

C)3.4 m/s D)2.9 m/s.​

Answer 1

Answer:

(D) 2.9 m/s

Explanation:

1) Let the take off point be 'C' . B

At 10:00 AM , balloon is at point 'D' and at 10:02 AM, balloon is at point 'A'.

2) In ΔABC,

$\angle ABC=60^{\circ}$

BC=300 m

$tan(60^{\circ})=\frac{AC}{BC}\\ \\=> \sqrt{3}=\frac{AC}{300}\\ \\=>AC=300\sqrt{3}m$

3) In Δ BCD ,

$\angle CBD=30^{\circ}$

$tan(30^{\circ})=\frac{CD}{BC}\\ \\=>\frac{1}{\sqrt{3}}=\frac{CD}{300}\\ \\=>CD=\frac{300}{\sqrt{3}}m$

4)  AD = AC - CD

 $=300\sqrt{3}-\frac{300}{\sqrt{3}}\\ \\=\frac{600}{\sqrt{3}} m$

5) It took 2 minutes to move from D to A.

=> 60*2=120s to move from D to A.

Hence, Speed of Balloon is

$\frac{600}{\sqrt{3}*120}=\frac{5}{\sqrt{3}} m = 2.9 m/s$