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Answer :
Sure! Let's solve each part of the problem step by step.
### Part a: How long will it take for the rocket to return to the ground?
The height [tex]\( h(t) \)[/tex] of the rocket is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
To find the time when the rocket returns to the ground, we need to set the height [tex]\( h(t) \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ -16t + 128 = 0 \][/tex]
Solving the second equation:
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{128}{16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket returns to the ground after [tex]\( t = 0 \)[/tex] seconds (the initial launch time) and [tex]\( t = 8 \)[/tex] seconds.
### Part b: After how many seconds will the rocket be 112 feet above the ground?
We need to set the height [tex]\( h(t) \)[/tex] to 112 feet and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
Rearranging the equation:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where:
[tex]\[ a = -16 \][/tex]
[tex]\[ b = 128 \][/tex]
[tex]\[ c = -112 \][/tex]
Let's solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4(-16)(-112)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-128 \pm \sqrt{16384 - 7168}}{-32} \][/tex]
[tex]\[ t = \frac{-128 \pm \sqrt{9216}}{-32} \][/tex]
[tex]\[ t = \frac{-128 \pm 96}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t = \frac{-128 + 96}{-32} = 1 \][/tex]
[tex]\[ t = \frac{-128 - 96}{-32} = 7 \][/tex]
So, the rocket will be 112 feet above the ground after [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
To summarize:
a. The rocket returns to the ground after 0 seconds and 8 seconds.
b. The rocket will be 112 feet above the ground after 1 second and 7 seconds.
### Part a: How long will it take for the rocket to return to the ground?
The height [tex]\( h(t) \)[/tex] of the rocket is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
To find the time when the rocket returns to the ground, we need to set the height [tex]\( h(t) \)[/tex] to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ -16t + 128 = 0 \][/tex]
Solving the second equation:
[tex]\[ -16t + 128 = 0 \][/tex]
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{128}{16} \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket returns to the ground after [tex]\( t = 0 \)[/tex] seconds (the initial launch time) and [tex]\( t = 8 \)[/tex] seconds.
### Part b: After how many seconds will the rocket be 112 feet above the ground?
We need to set the height [tex]\( h(t) \)[/tex] to 112 feet and solve for [tex]\( t \)[/tex]:
[tex]\[ -16t^2 + 128t = 112 \][/tex]
Rearranging the equation:
[tex]\[ -16t^2 + 128t - 112 = 0 \][/tex]
We solve this quadratic equation using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where:
[tex]\[ a = -16 \][/tex]
[tex]\[ b = 128 \][/tex]
[tex]\[ c = -112 \][/tex]
Let's solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4(-16)(-112)}}{2(-16)} \][/tex]
[tex]\[ t = \frac{-128 \pm \sqrt{16384 - 7168}}{-32} \][/tex]
[tex]\[ t = \frac{-128 \pm \sqrt{9216}}{-32} \][/tex]
[tex]\[ t = \frac{-128 \pm 96}{-32} \][/tex]
This gives us two solutions:
[tex]\[ t = \frac{-128 + 96}{-32} = 1 \][/tex]
[tex]\[ t = \frac{-128 - 96}{-32} = 7 \][/tex]
So, the rocket will be 112 feet above the ground after [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
To summarize:
a. The rocket returns to the ground after 0 seconds and 8 seconds.
b. The rocket will be 112 feet above the ground after 1 second and 7 seconds.
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