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Answer :
Let's solve the given problem step by step.
The height [tex]\( h \)[/tex] of the rocket after [tex]\( t \)[/tex] seconds is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
### 1. How long will it take for the rocket to return to the ground?
To find when the rocket returns to the ground, we need to determine when the height [tex]\( h \)[/tex] is equal to 0:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
First, factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
and
[tex]\[ -16t + 128 = 0 \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{-128}{-16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket will return to the ground at [tex]\( t = 8 \)[/tex] seconds. The [tex]\( t = 0 \)[/tex] solution represents the initial launch time.
### 2. After how many seconds will the rocket be 112 feet above the ground?
To find when the rocket is 112 feet above the ground, we set [tex]\( h(t) \)[/tex] to 112:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
Rearrange the equation to set it to 0:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
Let's solve this quadratic equation. This can be factored, though using the quadratic formula is straightforward:
The quadratic formula is:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -16 \)[/tex], [tex]\( b = 128 \)[/tex], and [tex]\( c = -112 \)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 128^2 - 4(-16)(-112) = 16384 - 7168 = 9216 \][/tex]
Now, find the square root of the discriminant:
[tex]\[ \sqrt{9216} = 96 \][/tex]
So, the solutions for [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{-128 \pm 96}{2(-16)} \][/tex]
Let's calculate each solution:
[tex]\[ t = \frac{128 + 96}{32} = \frac{224}{32} = 7 \][/tex]
and
[tex]\[ t = \frac{128 - 96}{32} = \frac{32}{32} = 1 \][/tex]
Therefore, the rocket will be 112 feet above the ground at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
### Summary
- The rocket will return to the ground at [tex]\( t = 8 \)[/tex] seconds.
- The rocket will be 112 feet above the ground at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
The height [tex]\( h \)[/tex] of the rocket after [tex]\( t \)[/tex] seconds is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
### 1. How long will it take for the rocket to return to the ground?
To find when the rocket returns to the ground, we need to determine when the height [tex]\( h \)[/tex] is equal to 0:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
First, factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
and
[tex]\[ -16t + 128 = 0 \][/tex]
Solve for [tex]\( t \)[/tex]:
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{-128}{-16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket will return to the ground at [tex]\( t = 8 \)[/tex] seconds. The [tex]\( t = 0 \)[/tex] solution represents the initial launch time.
### 2. After how many seconds will the rocket be 112 feet above the ground?
To find when the rocket is 112 feet above the ground, we set [tex]\( h(t) \)[/tex] to 112:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
Rearrange the equation to set it to 0:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
Let's solve this quadratic equation. This can be factored, though using the quadratic formula is straightforward:
The quadratic formula is:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -16 \)[/tex], [tex]\( b = 128 \)[/tex], and [tex]\( c = -112 \)[/tex].
Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac = 128^2 - 4(-16)(-112) = 16384 - 7168 = 9216 \][/tex]
Now, find the square root of the discriminant:
[tex]\[ \sqrt{9216} = 96 \][/tex]
So, the solutions for [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{-128 \pm 96}{2(-16)} \][/tex]
Let's calculate each solution:
[tex]\[ t = \frac{128 + 96}{32} = \frac{224}{32} = 7 \][/tex]
and
[tex]\[ t = \frac{128 - 96}{32} = \frac{32}{32} = 1 \][/tex]
Therefore, the rocket will be 112 feet above the ground at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
### Summary
- The rocket will return to the ground at [tex]\( t = 8 \)[/tex] seconds.
- The rocket will be 112 feet above the ground at [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
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