You throw a baseball directly upward at time=0 at an initial speed of 14.5 m/s. what is the maximum height the ball reaches above where it leaves your hand?
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Expert's answer
2012-10-05T07:56:17-0400
At the highest point: V = 0, V = V0 - g * t, so t = V0 / g The maximum height the ball reaches is: h = V0 * t - g * t^2 / 2 h = V0 * V0 / g - g * (V0 / g)^2 / 2 = V0^2 / (2 * g) h = 14.5^2 / (2 * 9.81) = 10.7 m
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