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Answer :
We start with the height function for the rocket:
$$
h(t) = -16t^2 + 32t + 48,
$$
where $t$ is the time in seconds and $h(t)$ is the height in feet.
### Part (a): Maximum Height of the Rocket
For a quadratic function of the form
$$
h(t) = at^2 + bt + c,
$$
the maximum (or minimum) occurs at the vertex. The time at which the vertex occurs is given by:
$$
t = -\frac{b}{2a}.
$$
Here, $a = -16$ and $b = 32$. Thus,
$$
t = -\frac{32}{2(-16)} = 1 \text{ second}.
$$
To find the maximum height, we substitute $t = 1$ back into the height function:
$$
h(1) = -16(1)^2 + 32(1) + 48.
$$
Calculating, we get
$$
h(1) = -16 + 32 + 48 = 64 \text{ feet}.
$$
So, the maximum height of the rocket is $\boxed{64 \text{ feet}}$, and it is reached at $1$ second.
### Part (b): Time When the Rocket Reaches the Ground
The rocket reaches the ground when $h(t) = 0$. Thus, we set the height function equal to zero:
$$
-16t^2 + 32t + 48 = 0.
$$
To solve this quadratic equation, we use the quadratic formula:
$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$
Substituting in the values $a = -16$, $b = 32$, and $c = 48$, we first calculate the discriminant:
$$
\Delta = b^2 - 4ac = 32^2 - 4(-16)(48).
$$
Simplify the discriminant:
$$
\Delta = 1024 + 3072 = 4096.
$$
Taking the square root:
$$
\sqrt{\Delta} = 64.
$$
Now substitute into the quadratic formula:
$$
t = \frac{-32 \pm 64}{2(-16)}.
$$
This gives two potential solutions:
1. Using the positive square root:
$$
t = \frac{-32 + 64}{-32} = \frac{32}{-32} = -1 \text{ second}
$$
2. Using the negative square root:
$$
t = \frac{-32 - 64}{-32} = \frac{-96}{-32} = 3 \text{ seconds}.
$$
Since time cannot be negative, we discard $t = -1$ second. Thus, the rocket reaches the ground at
$$
\boxed{3 \text{ seconds}}.
$$
### Final Answer
a. The maximum height of the rocket is $64$ feet (attained at $t = 1$ second).
b. The rocket reaches the ground at $3$ seconds.
$$
h(t) = -16t^2 + 32t + 48,
$$
where $t$ is the time in seconds and $h(t)$ is the height in feet.
### Part (a): Maximum Height of the Rocket
For a quadratic function of the form
$$
h(t) = at^2 + bt + c,
$$
the maximum (or minimum) occurs at the vertex. The time at which the vertex occurs is given by:
$$
t = -\frac{b}{2a}.
$$
Here, $a = -16$ and $b = 32$. Thus,
$$
t = -\frac{32}{2(-16)} = 1 \text{ second}.
$$
To find the maximum height, we substitute $t = 1$ back into the height function:
$$
h(1) = -16(1)^2 + 32(1) + 48.
$$
Calculating, we get
$$
h(1) = -16 + 32 + 48 = 64 \text{ feet}.
$$
So, the maximum height of the rocket is $\boxed{64 \text{ feet}}$, and it is reached at $1$ second.
### Part (b): Time When the Rocket Reaches the Ground
The rocket reaches the ground when $h(t) = 0$. Thus, we set the height function equal to zero:
$$
-16t^2 + 32t + 48 = 0.
$$
To solve this quadratic equation, we use the quadratic formula:
$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$
Substituting in the values $a = -16$, $b = 32$, and $c = 48$, we first calculate the discriminant:
$$
\Delta = b^2 - 4ac = 32^2 - 4(-16)(48).
$$
Simplify the discriminant:
$$
\Delta = 1024 + 3072 = 4096.
$$
Taking the square root:
$$
\sqrt{\Delta} = 64.
$$
Now substitute into the quadratic formula:
$$
t = \frac{-32 \pm 64}{2(-16)}.
$$
This gives two potential solutions:
1. Using the positive square root:
$$
t = \frac{-32 + 64}{-32} = \frac{32}{-32} = -1 \text{ second}
$$
2. Using the negative square root:
$$
t = \frac{-32 - 64}{-32} = \frac{-96}{-32} = 3 \text{ seconds}.
$$
Since time cannot be negative, we discard $t = -1$ second. Thus, the rocket reaches the ground at
$$
\boxed{3 \text{ seconds}}.
$$
### Final Answer
a. The maximum height of the rocket is $64$ feet (attained at $t = 1$ second).
b. The rocket reaches the ground at $3$ seconds.
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