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?

256ft

recall that the height h is given by

h(t) = 128t - 16t^2

That's just your good old friend, the parabola. Find its vertex -- that is the maximum height.

To find the maximum height reached by the firework, we can use the following kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity (0 ft/s when the firework reaches its maximum height)
u = initial velocity (128 ft/s)
a = acceleration (-16 ft/s^2, considering gravity)
s = displacement (unknown)

Substituting the given values into the equation:

0^2 = 128^2 + 2(-16)s

0 = 16384 - 32s

32s = 16384

s = 16384/32

s = 512 ft

So, the maximum height reached by the firework is 512 ft.

To find the maximum height reached by the firework, we can use the kinematic equation:

vf^2 = vi^2 + 2ad

where:
vf = final velocity (which is 0 at the maximum height)
vi = initial velocity (128 ft/s)
a = acceleration due to gravity (-16 ft/s^2)
d = displacement (the maximum height)

Rearranging the equation, we have:

d = (vf^2 - vi^2) / (2a)

Substituting the given values, we have:

d = (0^2 - 128^2) / (2(-16))

Simplifying the equation:

d = (-16384) / (-32)
d = 512 ft

Therefore, the maximum height reached by the firework is 512 ft.