Question

a baseball is thrown into the air from a height of 5 feet. the ball reaches a maximum height of 43.5 feet and spends a total of 3.2 seconds in the air. which equation models the height of the baseball? assume that acceleration due to gravity is - 16 ft/s².

$h(t)=16t^{2}+49.64t + 5$
$h(t)=-16t^{2}+5t + 49.64$
$h(t)=-16t^{2}+49.64t + 5$
$h(t)=16t^{2}+5t + 49.64$

Explanation:

Step1: Recall height - time formula

The general formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $h_{0}$ is the initial height, $v_{0}$ is the initial velocity, and $t$ is the time. The acceleration due to gravity $g = 32\ ft/s^{2}$, and the coefficient of $t^{2}$ is $-\frac{g}{2}=-16$. The initial height $h_{0}$ is given as 5 feet, so $h(t)=-16t^{2}+v_{0}t + 5$.

Step2: Use vertex formula to find initial velocity

The $t$ - value of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $t=-\frac{b}{2a}$. For $h(t)=-16t^{2}+v_{0}t + 5$, $a=-16$ and $b = v_{0}$. The ball reaches its maximum height at $t = 3.2$ seconds. Using the formula $t=-\frac{b}{2a}$, we have $3.2=-\frac{v_{0}}{2\times(-16)}$.
Solve for $v_{0}$:
\[

$$\begin{align*} 3.2&=\frac{v_{0}}{32}\\ v_{0}&=3.2\times32\\ v_{0}& = 49.64 \end{align*}$$

\]
So the height function is $h(t)=-16t^{2}+49.64t + 5$.

Answer:

$h(t)=-16t^{2}+49.64t + 5$