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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 :

MBERISSOGO

Answer:

magnitude = 13.02 m/s

direction = 86.9 degrees relative to ground.

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: [tex]\sqrt{13^2+0.7^2} \approx 13.02\,\,m/s[/tex]

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

[tex]\theta=arctan(\frac{13}{0.7}) =86.9^o[/tex]

DANIEGIRL77GO

Answer:

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.

Explanation:

Plato/Edmentum

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