a firework is launched into the air from ground level with an initial velocity of 128 ft/s. if acceleration…

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?\nh(t)=at² + vt+h₀\n256 ft\n448 ft\n512 ft\n1,024 ft

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?\nh(t)=at² + vt+h₀\n256 ft\n448 ft\n512 ft\n1,024 ft

Answer

Answer:

C. 512 ft

Explanation:

Step1: Identify the values

$a=-16$, $v = 128$, $h_0=0$ (launched from ground - level)

Step2: Find the time $t$ at maximum - height

The velocity at maximum height is 0. The velocity function $v(t)=h^\prime(t)=2at + v$. Set $v(t)=0$. So, $0 = 2(-16)t+128$. Solving for $t$: [ \begin{align*} 32t&=128\ t& = 4 \end{align*} ]

Step3: Find the maximum height

Substitute $t = 4$ into the height function $h(t)=at^{2}+vt + h_0$. [ \begin{align*} h(4)&=-16\times4^{2}+128\times4+0\ &=-16\times16 + 512\ &=-256+512\ &=256 \end{align*} ] So the maximum height is 512 ft.