physics

if a ball is thrown into the air at a speed of 22 m/s what is its position after 2.8 seconds?

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  1. Assuming it is thrown straight up, the altitude y after t seconds is

    y = 22t - (g/2)t^2

    where g = 9.8 m/s^2 is the acceleration of gravity

    I get 23.2 meters

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  2. Thank you so much!

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