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

a ball is thrown vertically upward. after t seconds, its height h (in feet) is given by the function h(t)=112t - 16t². what is the maximum height that the ball will reach? do not round your answer. height: feet
Answer
Answer:
196
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 $a=-16$, $b = 112$, $c = 0$.
Step2: Find the time $t$ at maximum - height
For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case $t$ - coordinate) of the vertex is given by $t=-\frac{b}{2a}$. Substituting $a=-16$ and $b = 112$ into the formula, we get $t=-\frac{112}{2\times(-16)}=\frac{112}{32}=\frac{7}{2}$.
Step3: Calculate the maximum height
Substitute $t = \frac{7}{2}$ into the height function $h(t)$. $h(\frac{7}{2})=112\times\frac{7}{2}-16\times(\frac{7}{2})^{2}$ $=56\times7-16\times\frac{49}{4}$ $=392 - 196$ $=196$.