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 $h(t)=112t - 16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 112$, 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 $t$ - coordinate) of the vertex is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 112$ into the formula: $t=-\frac{112}{2\times(-16)}=\frac{112}{32}=\frac{7}{2}$ seconds.

Step3: Find the maximum height

Substitute $t = \frac{7}{2}$ into the height - function $h(t)$: $h(\frac{7}{2})=112\times\frac{7}{2}-16\times(\frac{7}{2})^{2}$ $h(\frac{7}{2})=56\times7-16\times\frac{49}{4}$ $h(\frac{7}{2}) = 392-196$ $h(\frac{7}{2})=196$ feet.

Answer:

196