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Answer :
We are given the height function:
[tex]$$
f(t) = -16t^2 + 80t,
$$[/tex]
where [tex]$t$[/tex] is the time in seconds. To find the maximum height reached by the rocket, we locate the vertex of the parabola.
1. The formula for the time at which the maximum (vertex) occurs for a quadratic function [tex]$at^2 + bt + c$[/tex] is:
[tex]$$
t = \frac{-b}{2a}.
$$[/tex]
2. In our equation, [tex]$a = -16$[/tex] and [tex]$b = 80$[/tex]. Thus, we calculate:
[tex]$$
t = \frac{-80}{2(-16)} = \frac{-80}{-32} = 2.5 \text{ seconds}.
$$[/tex]
3. Next, we substitute [tex]$t = 2.5$[/tex] seconds into the height function to find the maximum height:
[tex]$$
\begin{aligned}
f(2.5) &= -16(2.5)^2 + 80(2.5) \\
&= -16(6.25) + 200 \\
&= -100 + 200 \\
&= 100 \text{ feet}.
\end{aligned}
$$[/tex]
Therefore, the maximum height of the rocket is:
[tex]$$
\boxed{100} \text{ feet}.
$$[/tex]
[tex]$$
f(t) = -16t^2 + 80t,
$$[/tex]
where [tex]$t$[/tex] is the time in seconds. To find the maximum height reached by the rocket, we locate the vertex of the parabola.
1. The formula for the time at which the maximum (vertex) occurs for a quadratic function [tex]$at^2 + bt + c$[/tex] is:
[tex]$$
t = \frac{-b}{2a}.
$$[/tex]
2. In our equation, [tex]$a = -16$[/tex] and [tex]$b = 80$[/tex]. Thus, we calculate:
[tex]$$
t = \frac{-80}{2(-16)} = \frac{-80}{-32} = 2.5 \text{ seconds}.
$$[/tex]
3. Next, we substitute [tex]$t = 2.5$[/tex] seconds into the height function to find the maximum height:
[tex]$$
\begin{aligned}
f(2.5) &= -16(2.5)^2 + 80(2.5) \\
&= -16(6.25) + 200 \\
&= -100 + 200 \\
&= 100 \text{ feet}.
\end{aligned}
$$[/tex]
Therefore, the maximum height of the rocket is:
[tex]$$
\boxed{100} \text{ feet}.
$$[/tex]
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