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Answer :
Sure! Let's go through the solution step by step.
### Part 1: Time for the Rocket to Return to the Ground
The height of the rocket [tex]\( h(t) \)[/tex] after [tex]\( t \)[/tex] seconds is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
To find when the rocket returns to the ground, we need to determine when [tex]\( h(t) = 0 \)[/tex]. So, we set the height equation to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We can factor out the common term, [tex]\( -16t \)[/tex]:
[tex]\[ -16t(t - 8) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( -16t = 0 \implies t = 0 \)[/tex]
2. [tex]\( t - 8 = 0 \implies t = 8 \)[/tex]
So, the rocket returns to the ground after [tex]\( 8 \)[/tex] seconds. Note that [tex]\( t = 0 \)[/tex] is the initial launch time.
### Part 2: Time for the Rocket to Be 112 Feet Above the Ground
Next, we need to determine the time(s) when the rocket is 112 feet above the ground. So, we set the height equation equal to 112 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
First, let's move all terms to one side of the equation:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
To simplify, we can divide the entire equation by -16:
[tex]\[ t^2 - 8t + 7 = 0 \][/tex]
Now, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 7 \)[/tex]:
[tex]\[ t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{64 - 28}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm 6}{2} \][/tex]
This gives us two solutions:
1. [tex]\( t = \frac{8 + 6}{2} = 7 \)[/tex]
2. [tex]\( t = \frac{8 - 6}{2} = 1 \)[/tex]
So, the rocket will be 112 feet above the ground at [tex]\( 1 \)[/tex] second and [tex]\( 7 \)[/tex] seconds after launch.
### Summary
1. The rocket returns to the ground after [tex]\( 8 \)[/tex] seconds.
2. The rocket is 112 feet above the ground at [tex]\( 1 \)[/tex] second and [tex]\( 7 \)[/tex] seconds after launch.
### Part 1: Time for the Rocket to Return to the Ground
The height of the rocket [tex]\( h(t) \)[/tex] after [tex]\( t \)[/tex] seconds is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
To find when the rocket returns to the ground, we need to determine when [tex]\( h(t) = 0 \)[/tex]. So, we set the height equation to zero and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We can factor out the common term, [tex]\( -16t \)[/tex]:
[tex]\[ -16t(t - 8) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( -16t = 0 \implies t = 0 \)[/tex]
2. [tex]\( t - 8 = 0 \implies t = 8 \)[/tex]
So, the rocket returns to the ground after [tex]\( 8 \)[/tex] seconds. Note that [tex]\( t = 0 \)[/tex] is the initial launch time.
### Part 2: Time for the Rocket to Be 112 Feet Above the Ground
Next, we need to determine the time(s) when the rocket is 112 feet above the ground. So, we set the height equation equal to 112 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
First, let's move all terms to one side of the equation:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
To simplify, we can divide the entire equation by -16:
[tex]\[ t^2 - 8t + 7 = 0 \][/tex]
Now, we use the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 7 \)[/tex]:
[tex]\[ t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{64 - 28}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ t = \frac{8 \pm 6}{2} \][/tex]
This gives us two solutions:
1. [tex]\( t = \frac{8 + 6}{2} = 7 \)[/tex]
2. [tex]\( t = \frac{8 - 6}{2} = 1 \)[/tex]
So, the rocket will be 112 feet above the ground at [tex]\( 1 \)[/tex] second and [tex]\( 7 \)[/tex] seconds after launch.
### Summary
1. The rocket returns to the ground after [tex]\( 8 \)[/tex] seconds.
2. The rocket is 112 feet above the ground at [tex]\( 1 \)[/tex] second and [tex]\( 7 \)[/tex] seconds after launch.
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