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)=48t - 16t². what is the maximum height that the ball will reach? do not round your answer. height: ft
Answer
Explanation:
Step1: Identify the function type
The height - function $h(t)=48t - 16t^{2}$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-16$, $b = 48$, and $c = 0$.
Step2: Find the time 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 $b = 48$ into the formula, we have $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}$ seconds.
Step3: Find the maximum height
Substitute $t = \frac{3}{2}$ into the height - function $h(t)$. $h(\frac{3}{2})=48\times\frac{3}{2}-16\times(\frac{3}{2})^{2}$ $h(\frac{3}{2})=72-16\times\frac{9}{4}$ $h(\frac{3}{2})=72 - 36$ $h(\frac{3}{2}) = 36$ feet.
Answer:
36