Question

You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk moving directly away from the balloon in a direction due north. The speed of the hawk relative to you is 2.0 m/s. What are the magnitude and direction of the hawk’s velocity relative to the ground? Express the directional angle rel- ative to due east.

Answer 1

6.32 m/s 18.43° northeast

Step-by-step explanation:

We express the velocity of hawk as:

`v_(Hawk)=v_(balloon)+v_(HawkRelativetoBalloon)=6 x+2 y`

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

`|v_(hawk)|=\sqrt[]{6^2+2^2}=√(40)`≈
`6.32 m/s`

And the angle with respect to the east should be with:

`arctan((2)/(6) )=18.43 \°`