Question

A hot air balloon rises at a constant speed of 13 meters/second relative to the air. There is a wind blowing eastwards at a speed of 0.7 meters/second relative to the ground. What is the magnitude and direction of the balloon’s velocity relative to the ground? Use the Pythagorean theorem to verify the answer.

Answer 1

Answer:

13.02 m/s the velocity and 86.92 degrees the direction relative to ground

Explanation:

We need to add velocities in vector addition to find the resultant velocity "$v$" of the balloon (the 13 m/s and the 0.7 m/s).

The velocities are at 90 degrees from each other (one pointing up and the other to the East). Notice from the attached image that the resultant velocity vector (picture in red) is actually the hypotenuse of a right angle triangle.

So we use Pythagoras to find the length (magnitude) of the resultant velocity vector:

$v = \sqrt{13^2 +0.7^2} = 13.018832...$

we can round the answer to 13.02 m/s

Now we need to find the angle that this new vector makes with the ground by using the definition of tangent of an angle that relates the two quantities that we just added:

$tan (\theta) = 13/0.7 \\ \theta = arctan (\frac{13}{07} = 86.91781918...$

So we round it to 86.92 degrees