A firework 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

The height is h(t) = −16t² + 128t, since the firework starts at ground level. Its maximum occurs at the vertex, at t = −128 ÷ (2 × −16) = 4 seconds. Substituting gives h(4) = −16(4²) + 128(4) = 256 feet, so the maximum height is 256 ft.

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