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 ]

a. 256 ft

b. 448 ft

c. 512 ft

d. 1,024 ft

Given that the equation to find the height of the firework is

h (t) = at² + vt + h₀

with a = - 16 ft/s² and v = 128 ft/s. In addition, since the firework starts from the ground, then the initial height, h₀, is equal to 0. Substituting these values, we have

h (t) = - 16t² + 128t + 0 = - 16t² + 128t

Seeing that h (t) is a quadratic function, then it forms a parabola. To find its maximum height, we can compute for the parabola's vertex.

To find the vertex's x-coordinate, we can use

t = - b/2a = (-128) / (2 · - 16) = - 128/-32 = 4

Since, it takes 4 seconds for the firework to reach its maximum height, then the maximum height it reaches is equal to h (4). Hence, we have

h (4) = - 16 (4) ² + 128 (4) = - 16 (16) + 512 = 256

Hence, the highest that the firework can reach is equal to 256 ft.

Answer: A. 256 ft