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: feet
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 $t$ at maximum - height
For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate (in our case $t$) of the vertex is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 48$ into the formula: $t=-\frac{48}{2\times(-16)}=\frac{48}{32}=\frac{3}{2}$ seconds.
Step3: Find the maximum height
Substitute $t = \frac{3}{2}$ into the height function $h(t)=48t-16t^{2}$: $h(\frac{3}{2})=48\times\frac{3}{2}-16\times(\frac{3}{2})^{2}$ $h(\frac{3}{2})=72 - 16\times\frac{9}{4}$ $h(\frac{3}{2})=72-36$ $h(\frac{3}{2}) = 36$ feet.
Answer:
36