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Answer :
Sure! Let's break down the solution step by step for each part of the question.
### a. Set up an equation to determine how long the rocket is in the air.
To find out how long the rocket is in the air, we set the height equation equal to zero because the rocket will be on the ground when its height is zero:
[tex]\[ h(t) = -16t^2 + 128t = 0 \][/tex]
We factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( t = 0 \)[/tex] (the rocket is at the ground initially)
2. Solve [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket is in the air for 8 seconds before it hits the ground.
### b. When does the rocket hit the ground?
From our previous calculations, the rocket hits the ground at [tex]\( t = 8 \)[/tex] seconds.
### c. Sketch a picture to model the data.
While I can't draw a picture here, you can imagine the path of the rocket as a parabola opening downwards. The curve starts at the origin (0, 0), reaches a maximum height at the vertex, and then returns to the x-axis at (8, 0), where the rocket hits the ground.
### d. At what time will the rocket reach its maximum height? What is the maximum height?
The maximum height of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is found at the vertex, which occurs at:
[tex]\[ t = -\frac{b}{2a} \][/tex]
In our equation, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex]:
[tex]\[ t = -\frac{128}{2 \times -16} = 4 \][/tex]
This means the rocket reaches its maximum height at [tex]\( t = 4 \)[/tex] seconds. Now, we find the maximum height by plugging [tex]\( t = 4 \)[/tex] back into the height equation:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16(16) + 128(4) \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
So, the maximum height reached by the rocket is 256 feet.
### a. Set up an equation to determine how long the rocket is in the air.
To find out how long the rocket is in the air, we set the height equation equal to zero because the rocket will be on the ground when its height is zero:
[tex]\[ h(t) = -16t^2 + 128t = 0 \][/tex]
We factor out the common term [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( t = 0 \)[/tex] (the rocket is at the ground initially)
2. Solve [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = 8 \][/tex]
So, the rocket is in the air for 8 seconds before it hits the ground.
### b. When does the rocket hit the ground?
From our previous calculations, the rocket hits the ground at [tex]\( t = 8 \)[/tex] seconds.
### c. Sketch a picture to model the data.
While I can't draw a picture here, you can imagine the path of the rocket as a parabola opening downwards. The curve starts at the origin (0, 0), reaches a maximum height at the vertex, and then returns to the x-axis at (8, 0), where the rocket hits the ground.
### d. At what time will the rocket reach its maximum height? What is the maximum height?
The maximum height of a quadratic equation [tex]\( ax^2 + bx + c \)[/tex] is found at the vertex, which occurs at:
[tex]\[ t = -\frac{b}{2a} \][/tex]
In our equation, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex]:
[tex]\[ t = -\frac{128}{2 \times -16} = 4 \][/tex]
This means the rocket reaches its maximum height at [tex]\( t = 4 \)[/tex] seconds. Now, we find the maximum height by plugging [tex]\( t = 4 \)[/tex] back into the height equation:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16(16) + 128(4) \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
So, the maximum height reached by the rocket is 256 feet.
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