this is in addition to my previous question (the particle undergoing an acceleration of 2.5 m/s^2 to the right and 3.2 m/s^2 up. what is its speed after 5.9s?)

What is the direction with respect to the horizontal at this time? Answer between -180 degrees and 180 degrees. Anwer in units of degrees.

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  1. tan^-1(Vz/Vx) = tan^-1(az/ax)
    = tan^-1(5.9/3.2) = 61.5 degrees

    The angle will be independent of time.

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  2. thanks!

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