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Suppose a ball is thrown upward at a velocity of 44ft/sec from a cliff 200 feet above a dry riverbed. Predict the height of the ball after 3 seconds, and then determine the time that the ball will hit the riverbed.
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y = (1/2)*g*t^2 +v0*t + y0 so your y0 = 200 feet. for the first part of the question, the time = 3 and the initial velocity = 44 ft/s gravity = -32 ft/s^2 so you have y = (1/2)*(-32)*3^2 +44*3 + 200 therefore y = 188 (check my math with a calculator, I did it in my head.)
\[y-y_0 =v_0t - {1 \over 2} g t^2\]
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