Question

the height of a softball, in feet, is modeled by the function h given by h(t)=3 + 55t - 16t², where t is the time, in seconds, after softball is hit. a graph of the function is shown. type your answers in the boxes. a. about when does the softball reach its maximum height? b. about how high is the maximum height of the softball?

Explanation:

Step1: Identify the function type

The height - function $h(t)=3 + 55t-16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 55$, and $c = 3$.

Step2: Find the time of maximum height

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case, the time $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 55$ into the formula:
\[t=-\frac{55}{2\times(-16)}=\frac{55}{32}\approx1.72\] seconds.

Step3: Find the maximum height

Substitute $t=\frac{55}{32}$ into the function $h(t)=3 + 55t-16t^{2}$:
\[h(\frac{55}{32})=3+55\times\frac{55}{32}-16\times(\frac{55}{32})^{2}\]
\[=3+\frac{3025}{32}-16\times\frac{3025}{1024}\]
\[=3+\frac{3025}{32}-\frac{3025}{64}\]
\[=\frac{192 + 6050-3025}{64}=\frac{3217}{64}\approx50.27\] feet.

Answer:

a. $1.72$ seconds
b. $50.27$ feet