A moderate wind accelerates a pebble over a horizontal xy plane with constant acceleration a=5m/(s^2)i + 7m/(s^2)j

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A moderate wind accelerates a pebble over a horizontal xy plane with constant acceleration a=5m/(s^2)i + 7m/(s^2)j. AT time t=0 the velocity is 4.0 m/s. What is the velocity when it has been displaced 12m parallel to the x axis? found the answer, my question is why doesn’t the 4m/s starting velocity apply to the vertical motion as well?

Answer:

Direction of initial velocity is not given, so we can assume that the direction is the same as the direction of wind.

Direction of the wind is:

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