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A bsaeball is hit straight up at an initial velocity of 30 m/s. If the ball has a negative acceration of about 10 m/s^2, how long does the ball take to reach teh top of it's path?  

jessica552

by jessica552 at September 13, 2010

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At its height, its velocity is 0. So you have (30m/s)/(10m/s^s) = 3 m/m * s^2/s = 3s

John_Doe John_Doe September 14, 2010

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