Question

we are standing on the top of a 720 feet tall building and launch a small object upward. the objects vertical position, measured in feet, after t seconds is h(t)= - 16t^2+192t + 720. what is the highest point that the object reaches? feet

Explanation:

Step1: Identify the function type

The height - function $h(t)=-16t^{2}+192t + 720$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 192$, and $c = 720$.

Step2: Find the time $t$ at which the object reaches its maximum height

The time $t$ at which a quadratic function $y = ax^{2}+bx + c$ reaches its maximum (when $a<0$) is given by the formula $t=-\frac{b}{2a}$.
Substitute $a=-16$ and $b = 192$ into the formula:
$t=-\frac{192}{2\times(-16)}=\frac{-192}{-32}=6$.

Step3: Find the maximum height

Substitute $t = 6$ into the height - function $h(t)=-16t^{2}+192t + 720$.
$h(6)=-16\times6^{2}+192\times6 + 720$.
First, calculate $-16\times6^{2}=-16\times36=-576$.
Then, calculate $192\times6 = 1152$.
$h(6)=-576+1152 + 720$.
$h(6)=1396$.

Answer:

$1396$