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A ball it hit straight up in the air with an initial velocity of 12.9 m/s. At what time does it cross the height of 5 m going up?
3 years ago

Answers

Anonymous
5 = 12.9 t - 4.9 t^2
4.9 t^2 - 12.9 t + 5 = 0
t = 0.472 going up and 2.16 coming down
3 years ago

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