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An arrow is shot straight up in the air at an initial speed of 58.0 m/s. After how much time is the arrow heading downward at a speed of 2.00 m/s?

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Answer:

3.50 s

Step-by-step explanation:

Vertical speed = vo - 1/2 a t^2 vo = original velocity = 58 m/s

-2 m/s = 58 - 1/2 (9.81) t^2

t = 3.497 s = 3.50 s ( for three significant digits)

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