Question

A hot air balloon is released into the air. During its straight ascent, the angle of elevation was 15° and, 3 minutes later, the angle of elevation increased 20°. How fast is the balloon traveling, in km/h, if the angle measurements were taken 300m away from the launch site?

Answer 1

Answer:

The speed of the balloon is 0.16 m/s.

Step-by-step explanation:

CD = 300 m

Let AD = x

AB = y

time, t = 3 min

Triangle, ADC

$tan 15 = \frac{AD}{BC}\\\\0.27 \times 300 = x \\\\x = 80.4 m$

Triangle, BCD

$tan 20 = \frac{BD}{BC}\\\\0.36 \times 300 = x + y \\\\x + y = 109.2 m$

So, y = 109.2 - 80.4 = 28.8 m

Speed = 28.8/180 = 0.16 m/s