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

According to the Pythagorean theorem, the magnitude of the balloon’s velocity relative to the ground is = 13.02 ≈13.0 meters/seconds. The direction of the balloon relative to the ground is 3° northeast.

Step-by-step explanation:

Plato/Edmentum

Answer 2

magnitude = 13.02 m/s

direction = 86.9 degrees relative to ground.

Step-by-step explanation:

We need to compose the velocities in perpendicular directions using the Pythagoras theorem to find the magnitude of the composition:

magnitude of new velocity:
`√(13^2+0.7^2) \approx 13.02\,\,m/s`

The direction will be given by the angle relative to ground using the arctan function:

`\theta=arctan((13)/(0.7)) =86.9^o`