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 balloon is 0.16 m/s.

Step-by-step explanation:

BC = 300 m

AD = x

AB = y

time, t = 3 min = 3 x 60 = 180 s

By the triangle, ADC,

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

By the triangle, BDC

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

So, y = 109.2 - 80.4 = 28.8 m

Speed of balloon = 28.8/180 = 0.16 m/s