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Answer :
We are given the height function of the rocket:
$$
f(t) = -16t^2 + 80t.
$$
**Step 1: Determine the time of maximum height.**
A quadratic function in the form
$$
f(t) = at^2 + bt + c
$$
achieves its maximum (or minimum) at the vertex. The time coordinate of the vertex is
$$
t = -\frac{b}{2a}.
$$
In our function, $a = -16$ and $b = 80$. Substituting these values gives:
$$
t = -\frac{80}{2(-16)} = \frac{80}{32} = 2.5 \text{ seconds}.
$$
**Step 2: Find the maximum height.**
Substitute $t = 2.5$ back into the height function:
$$
f(2.5) = -16(2.5)^2 + 80(2.5).
$$
Calculate $2.5^2$:
$$
(2.5)^2 = 6.25.
$$
Then, compute the terms:
$$
-16 \times 6.25 = -100,
$$
$$
80 \times 2.5 = 200.
$$
Add these together:
$$
f(2.5) = -100 + 200 = 100 \text{ feet}.
$$
**Conclusion:**
The maximum height reached by the rocket is
$$
\boxed{100} \text{ feet}.
$$
$$
f(t) = -16t^2 + 80t.
$$
**Step 1: Determine the time of maximum height.**
A quadratic function in the form
$$
f(t) = at^2 + bt + c
$$
achieves its maximum (or minimum) at the vertex. The time coordinate of the vertex is
$$
t = -\frac{b}{2a}.
$$
In our function, $a = -16$ and $b = 80$. Substituting these values gives:
$$
t = -\frac{80}{2(-16)} = \frac{80}{32} = 2.5 \text{ seconds}.
$$
**Step 2: Find the maximum height.**
Substitute $t = 2.5$ back into the height function:
$$
f(2.5) = -16(2.5)^2 + 80(2.5).
$$
Calculate $2.5^2$:
$$
(2.5)^2 = 6.25.
$$
Then, compute the terms:
$$
-16 \times 6.25 = -100,
$$
$$
80 \times 2.5 = 200.
$$
Add these together:
$$
f(2.5) = -100 + 200 = 100 \text{ feet}.
$$
**Conclusion:**
The maximum height reached by the rocket is
$$
\boxed{100} \text{ feet}.
$$
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