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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.

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yesquitpipe

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

[tex]v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y[/tex]

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

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

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

[tex]arctan(\frac{2}{6} )=18.43 \°[/tex]

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