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Matching heights A stone is thrown from the edge of a bridge that is 48 ft above the ground with an initial velocity of 32 ft/s. The height of this stone above the ground t seconds after it is thrown is f(t) = −16t2 + 32t + 48. If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t) = −16t2 + ν0 t, where ν0 is the initial velocity of the second stone. Determine the value of ν0 such that both stones reach the same high point.
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Calculus: Early Transcendentals (3rd Edition)
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