Question

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?
$h(t) = at^2 + vt + h_0$
○ 256 ft
○ 448 ft
○ 512 ft
○ 1,024 ft

Explanation:

Step1: Define height function parameters

$a = -16$, $v = 128$, $h_0 = 0$, so $h(t) = -16t^2 + 128t$

Step2: Find time of maximum height

For quadratic $ax^2+bx+c$, $t = -\frac{b}{2a} = -\frac{128}{2(-16)} = 4$

Step3: Calculate max height at t=4

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

Answer:

256 ft