a ball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function…

Explanation:

Step1: Identify the function type

The height - function $h(t)=104t - 16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 104$, and $c = 0$.

Step2: Find the time $t$ at which the maximum occurs

For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case, the $t$ - coordinate) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 104$ into the formula, we have $t=-\frac{104}{2\times(-16)}=\frac{104}{32}=\frac{13}{4}$.

Step3: Find the maximum height

Substitute $t = \frac{13}{4}$ into the height - function $h(t)$. [ \begin{align*} h\left(\frac{13}{4}\right)&=104\times\frac{13}{4}-16\times\left(\frac{13}{4}\right)^{2}\ &=26\times13-16\times\frac{169}{16}\ &=338 - 169\ &=169 \end{align*} ]

Answer:

169