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A stone thrown from the ground reaches a height of 64ft in 1 second. Determine the maximum height of the stone.?

Calculus problem. I can not understand

h =- 16t ^ 2 + v0t where v0 is the initial velocity. To see this: the acceleration a (t) = -32 m / s ^ 2 (negative because we are considering as the address positive and gravity pulls it down.) for v (t) = Integral (a (t), t) =- 32t + V0 and position (or height) is: h (t) = Integral (v (t) , t) =- 16t ^ 2 + v0t + h0 (where is the initial height h0) h0 = 0, since the object is launched from the ground (something hard to do, if you think about it.) Using h = 64 at t = 1: 64 = -16 + V0 for v0 = 80 m / sec. So v = 32t + 80 and the maximum height occurs when v = 0 or T = 80/32 = 2.5 sec. To find the maximum height to find h (2.5) = -16 (2.5) ^ 2 + 80 * 2.5 = 100 feet.

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