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Answer :
We are given the height function
[tex]$$
h(t) = -16t^2 + 32t + 48
$$[/tex]
for the rocket, where [tex]$t$[/tex] is time in seconds and [tex]$h(t)$[/tex] is the height in feet.
Step 1. Determine the time when the rocket reaches its maximum height.
For a quadratic function of the form
[tex]$$
ax^2 + bx + c,
$$[/tex]
the vertex (which gives the maximum for a downward opening parabola, [tex]$a < 0$[/tex]) occurs at
[tex]$$
t = -\frac{b}{2a}.
$$[/tex]
Here, [tex]$a = -16$[/tex] and [tex]$b = 32$[/tex]. Thus,
[tex]$$
t = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1.
$$[/tex]
This means the maximum height occurs at [tex]$t = 1$[/tex] second.
Step 2. Find the maximum height.
Substitute [tex]$t = 1$[/tex] into the height function:
[tex]$$
h(1) = -16(1)^2 + 32(1) + 48 = -16 + 32 + 48 = 64.
$$[/tex]
So, the maximum height is [tex]$\boxed{64}$[/tex] feet.
Step 3. Determine the time when the rocket reaches the ground.
The rocket reaches the ground when [tex]$h(t) = 0$[/tex]. Set the height equation equal to zero:
[tex]$$
-16t^2 + 32t + 48 = 0.
$$[/tex]
It can be easier to work with positive leading coefficients. Multiply the entire equation by [tex]$-1$[/tex]:
[tex]$$
16t^2 - 32t - 48 = 0.
$$[/tex]
Step 4. Solve the quadratic equation using the quadratic formula:
[tex]$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$[/tex]
In the equation [tex]$16t^2 - 32t - 48 = 0$[/tex], we identify [tex]$a = 16$[/tex], [tex]$b = -32$[/tex], and [tex]$c = -48$[/tex]. Compute the discriminant:
[tex]$$
\Delta = (-32)^2 - 4(16)(-48) = 1024 + 3072 = 4096.
$$[/tex]
Taking the square root gives:
[tex]$$
\sqrt{\Delta} = 64.
$$[/tex]
Now, solve for [tex]$t$[/tex]:
[tex]$$
t = \frac{-(-32) \pm 64}{2 \times 16} = \frac{32 \pm 64}{32}.
$$[/tex]
This yields two solutions:
1. When choosing the positive square root:
[tex]$$
t = \frac{32 + 64}{32} = \frac{96}{32} = 3.
$$[/tex]
2. When choosing the negative square root:
[tex]$$
t = \frac{32 - 64}{32} = \frac{-32}{32} = -1.
$$[/tex]
Since time cannot be negative, we discard [tex]$t = -1$[/tex]. Thus, the rocket reaches the ground at [tex]$\boxed{3}$[/tex] seconds.
Final Answer:
a. The maximum height of the rocket is [tex]$64$[/tex] feet.
b. The rocket reaches the ground after [tex]$3$[/tex] seconds.
[tex]$$
h(t) = -16t^2 + 32t + 48
$$[/tex]
for the rocket, where [tex]$t$[/tex] is time in seconds and [tex]$h(t)$[/tex] is the height in feet.
Step 1. Determine the time when the rocket reaches its maximum height.
For a quadratic function of the form
[tex]$$
ax^2 + bx + c,
$$[/tex]
the vertex (which gives the maximum for a downward opening parabola, [tex]$a < 0$[/tex]) occurs at
[tex]$$
t = -\frac{b}{2a}.
$$[/tex]
Here, [tex]$a = -16$[/tex] and [tex]$b = 32$[/tex]. Thus,
[tex]$$
t = -\frac{32}{2(-16)} = -\frac{32}{-32} = 1.
$$[/tex]
This means the maximum height occurs at [tex]$t = 1$[/tex] second.
Step 2. Find the maximum height.
Substitute [tex]$t = 1$[/tex] into the height function:
[tex]$$
h(1) = -16(1)^2 + 32(1) + 48 = -16 + 32 + 48 = 64.
$$[/tex]
So, the maximum height is [tex]$\boxed{64}$[/tex] feet.
Step 3. Determine the time when the rocket reaches the ground.
The rocket reaches the ground when [tex]$h(t) = 0$[/tex]. Set the height equation equal to zero:
[tex]$$
-16t^2 + 32t + 48 = 0.
$$[/tex]
It can be easier to work with positive leading coefficients. Multiply the entire equation by [tex]$-1$[/tex]:
[tex]$$
16t^2 - 32t - 48 = 0.
$$[/tex]
Step 4. Solve the quadratic equation using the quadratic formula:
[tex]$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
$$[/tex]
In the equation [tex]$16t^2 - 32t - 48 = 0$[/tex], we identify [tex]$a = 16$[/tex], [tex]$b = -32$[/tex], and [tex]$c = -48$[/tex]. Compute the discriminant:
[tex]$$
\Delta = (-32)^2 - 4(16)(-48) = 1024 + 3072 = 4096.
$$[/tex]
Taking the square root gives:
[tex]$$
\sqrt{\Delta} = 64.
$$[/tex]
Now, solve for [tex]$t$[/tex]:
[tex]$$
t = \frac{-(-32) \pm 64}{2 \times 16} = \frac{32 \pm 64}{32}.
$$[/tex]
This yields two solutions:
1. When choosing the positive square root:
[tex]$$
t = \frac{32 + 64}{32} = \frac{96}{32} = 3.
$$[/tex]
2. When choosing the negative square root:
[tex]$$
t = \frac{32 - 64}{32} = \frac{-32}{32} = -1.
$$[/tex]
Since time cannot be negative, we discard [tex]$t = -1$[/tex]. Thus, the rocket reaches the ground at [tex]$\boxed{3}$[/tex] seconds.
Final Answer:
a. The maximum height of the rocket is [tex]$64$[/tex] feet.
b. The rocket reaches the ground after [tex]$3$[/tex] seconds.
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