Question

1) A balloon is released that rises at a constant 8.0 m/s with respect to the air. A breeze is

blowing towards the East with a velocity of 4.0 m/s. What is the magnitude and direction of the
balloon's velocity?

Answer 1

`Magnitude = 8.94427 \approx 8.9 m`

`Angle = 63.43^\circ`

Step-by-step explanation:

Since the wind is blowing at 4 m/s east (+ x axis) and the balloon is rising at 8 m/s (+y axis), the magnitude and direction of the resultant vector can be found as

`Magnitude =√(4^2+8^2) = =√(16+64) = √(80) = 8.94427 \approx 8.9 m`

`Angle = \tan^(-1)\left({(8)/(4)\right) = \tan^(-1)(2) = 63.43^\circ`