Question

8.) an object is projected vertically upward from the top of a building with an initial velocity of 144 ft/sec. its distance s(t) in feet above the ground after t seconds is given by the equation s(t)=-16t² + 144t + 100.
a.) find its maximum distance above the ground.
b.) find the height of the building.

Explanation:

Step1: Identify the function type

The distance function $s(t)=-16t^{2}+144t + 100$ is a quadratic function in the form $y = ax^{2}+bx + c$ where $a=-16$, $b = 144$, $c = 100$.

Step2: Find the time at which maximum occurs

For a quadratic function $y=ax^{2}+bx + c$, the $t$-coordinate of the vertex (where maximum or minimum occurs) is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 144$, we have $t=-\frac{144}{2\times(-16)}=\frac{144}{32}=\frac{9}{2}=4.5$ seconds.

Step3: Find the maximum distance

Substitute $t = 4.5$ into the function $s(t)$.
\[

$$\begin{align*} s(4.5)&=-16\times(4.5)^{2}+144\times4.5 + 100\\ &=-16\times20.25+648+100\\ &=-324+648 + 100\\ &=424 \end{align*}$$

\]

Step4: Find the height of the building

The height of the building is the distance of the object from the ground at $t = 0$. Substitute $t=0$ into $s(t)$.
$s(0)=-16\times0^{2}+144\times0+100 = 100$ feet.

Answer:

a. 424 feet
b. 100 feet