A firework is launched into the air from ground level with an initial velocity of 128 ft/s If acceleration due to gravity is negative 16 the fraction with numerator ft and denominator s squared , what is the maximum height reached by the firework? h of t is equal to A t squared plus v t plus h sub 0

Using h(t) = at² + vt + h₀, the height is h(t) = −16t² + 128t because the firework starts at ground level. The maximum occurs at the vertex, at t = −v/(2a) = −128/(2 × −16) = 4 seconds. Substituting gives h(4) = −16(4²) + 128(4) = 256 feet, so the maximum height is 256 feet (choice C).

AI-written and checked against the verified answer.