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Answer :
Sure! Let's go through each part of the question step-by-step.
### Given:
The function for the height [tex]\( h(t) \)[/tex] of the rocket at time [tex]\( t \)[/tex] seconds is:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
### a. How long will it take for the rocket to return to the ground?
To find out when the rocket returns to the ground, we need to find when the height [tex]\( h(t) \)[/tex] is zero.
So, we set the equation to zero:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We can factor the equation:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
- [tex]\( t = 0 \)[/tex] (the time the rocket was launched)
- [tex]\( -16t + 128 = 0 \)[/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ 128 = 16t \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, it will take 8 seconds for the rocket to return to the ground.
### b. How long will it take the rocket to hit its maximum height?
The maximum height of a parabola given by [tex]\( h(t) = at^2 + bt + c \)[/tex] occurs at the vertex, which can be found using the formula [tex]\( t = -\frac{b}{2a} \)[/tex].
For our function:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 128 \)[/tex]
So we substitute these values into the formula:
[tex]\[ t = -\frac{128}{2 \times -16} \][/tex]
[tex]\[ t = \frac{128}{32} = 4 \][/tex]
Thus, the rocket reaches its maximum height at 4 seconds.
### c. What is the maximum height?
We find the maximum height by substituting the time at which the rocket reaches its maximum height back into the height function [tex]\( h(t) \)[/tex].
Plug [tex]\( t = 4 \)[/tex] into [tex]\( h(t) \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16 \cdot 16 + 128 \cdot 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Therefore, the maximum height is 256 feet.
These calculations show that the rocket takes 8 seconds to return to the ground, reaches its maximum height in 4 seconds, and the maximum height achieved is 256 feet.
### Given:
The function for the height [tex]\( h(t) \)[/tex] of the rocket at time [tex]\( t \)[/tex] seconds is:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
### a. How long will it take for the rocket to return to the ground?
To find out when the rocket returns to the ground, we need to find when the height [tex]\( h(t) \)[/tex] is zero.
So, we set the equation to zero:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
We can factor the equation:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
- [tex]\( t = 0 \)[/tex] (the time the rocket was launched)
- [tex]\( -16t + 128 = 0 \)[/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ 128 = 16t \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, it will take 8 seconds for the rocket to return to the ground.
### b. How long will it take the rocket to hit its maximum height?
The maximum height of a parabola given by [tex]\( h(t) = at^2 + bt + c \)[/tex] occurs at the vertex, which can be found using the formula [tex]\( t = -\frac{b}{2a} \)[/tex].
For our function:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 128 \)[/tex]
So we substitute these values into the formula:
[tex]\[ t = -\frac{128}{2 \times -16} \][/tex]
[tex]\[ t = \frac{128}{32} = 4 \][/tex]
Thus, the rocket reaches its maximum height at 4 seconds.
### c. What is the maximum height?
We find the maximum height by substituting the time at which the rocket reaches its maximum height back into the height function [tex]\( h(t) \)[/tex].
Plug [tex]\( t = 4 \)[/tex] into [tex]\( h(t) \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16 \cdot 16 + 128 \cdot 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \][/tex]
Therefore, the maximum height is 256 feet.
These calculations show that the rocket takes 8 seconds to return to the ground, reaches its maximum height in 4 seconds, and the maximum height achieved is 256 feet.
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