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 + v₀t, where v₀ is the initial velocity of the second stone. Determine the value of v₀ such that both stones reach the same high point.