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4. 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.

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4. 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) = -1612 + 321 + 48. If a second stone is thrown from the ground, then its height above the ground after t seconds is given by g(t) = -16t- + vot, where vo is the initial velocity of the second stone. Determine the value up so that both the stones reach the same high point

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