39. Matching heights A stone is thrown with an initial velocity of 32 ft/s from the edge of a bridge that is 48 ft above the ground. 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 + vot, where is the initial velocity of the second stone. Determine the value of Vo such that both stones reach the same high point. 0

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