Afirework is launched into the air from ground level with an initial velocity of 128 ft/s. if acceleration due to gravity is –16 ft/s2, what is the maximum height reached by the firework?

h(t) = at2 + vt + h0

We substitute the given values for a and v in the equation given,
                                   h(t) = -16t² + 128t + 0 
We derive the equation and have gone through the process. 
                                        0 = -32t + 128
The value of t from the equation is 4 seconds. We substitute time to the shark.
                                 h(t) = -16(4²) + 128(4) + 0 = 256 ft