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Answer :
Sure, let's go through each part of the question step-by-step.
### a. How long will it take for the rocket to return to the ground?
The height of the rocket when it returns to the ground is 0. This means we need to set the height equation [tex]\( h(t) = -16t^2 + 128t \)[/tex] to zero and solve for [tex]\( t \)[/tex].
1. Set the equation to zero:
[tex]\(-16t^2 + 128t = 0\)[/tex]
2. Factor out the common term, which is [tex]\( t \)[/tex]:
[tex]\( t(-16t + 128) = 0 \)[/tex]
3. Solve for [tex]\( t \)[/tex]:
From [tex]\( t(-16t + 128) = 0 \)[/tex], we get two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ -16t + 128 = 0 \][/tex]
4. Solve [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, the rocket returns to the ground after 8 seconds.
### b. After how many seconds will the rocket be 112 feet above the ground?
We need to set the height equation to 112 and solve for [tex]\( t \)[/tex]:
1. Set up the equation:
[tex]\(-16t^2 + 128t = 112\)[/tex]
2. Rearrange to form a quadratic equation:
[tex]\(-16t^2 + 128t - 112 = 0\)[/tex]
3. Solve the quadratic equation:
This equation leads to two possible solutions for [tex]\( t \)[/tex], which represent the times when the rocket is at 112 feet while going up and coming down.
The solutions are [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
### c. How long will it take the rocket to hit its maximum height?
The maximum height of a projectile (like our toy rocket) is reached at the vertex of the parabolic equation [tex]\( h(t) = -16t^2 + 128t \)[/tex].
1. The time to reach maximum height is given by the formula [tex]\(-b/(2a)\)[/tex] for an equation in the form [tex]\( at^2 + bt + c \)[/tex]. Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex].
2. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the formula:
[tex]\[ t = \frac{-128}{2 \times -16} = \frac{128}{32} = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### d. What is the maximum height?
To find the maximum height, substitute the time found in part c ([tex]\( t = 4 \)[/tex]) into the height equation:
1. Substitute [tex]\( t = 4 \)[/tex] into [tex]\( h(t) = -16t^2 + 128t \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
2. Calculate:
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 = 256 \][/tex]
The maximum height of the rocket is 256 feet.
I hope this helps you understand how to approach these types of problems! Let me know if you have any questions.
### a. How long will it take for the rocket to return to the ground?
The height of the rocket when it returns to the ground is 0. This means we need to set the height equation [tex]\( h(t) = -16t^2 + 128t \)[/tex] to zero and solve for [tex]\( t \)[/tex].
1. Set the equation to zero:
[tex]\(-16t^2 + 128t = 0\)[/tex]
2. Factor out the common term, which is [tex]\( t \)[/tex]:
[tex]\( t(-16t + 128) = 0 \)[/tex]
3. Solve for [tex]\( t \)[/tex]:
From [tex]\( t(-16t + 128) = 0 \)[/tex], we get two solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ -16t + 128 = 0 \][/tex]
4. Solve [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, the rocket returns to the ground after 8 seconds.
### b. After how many seconds will the rocket be 112 feet above the ground?
We need to set the height equation to 112 and solve for [tex]\( t \)[/tex]:
1. Set up the equation:
[tex]\(-16t^2 + 128t = 112\)[/tex]
2. Rearrange to form a quadratic equation:
[tex]\(-16t^2 + 128t - 112 = 0\)[/tex]
3. Solve the quadratic equation:
This equation leads to two possible solutions for [tex]\( t \)[/tex], which represent the times when the rocket is at 112 feet while going up and coming down.
The solutions are [tex]\( t = 1 \)[/tex] second and [tex]\( t = 7 \)[/tex] seconds.
### c. How long will it take the rocket to hit its maximum height?
The maximum height of a projectile (like our toy rocket) is reached at the vertex of the parabolic equation [tex]\( h(t) = -16t^2 + 128t \)[/tex].
1. The time to reach maximum height is given by the formula [tex]\(-b/(2a)\)[/tex] for an equation in the form [tex]\( at^2 + bt + c \)[/tex]. Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex].
2. Substitute [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the formula:
[tex]\[ t = \frac{-128}{2 \times -16} = \frac{128}{32} = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### d. What is the maximum height?
To find the maximum height, substitute the time found in part c ([tex]\( t = 4 \)[/tex]) into the height equation:
1. Substitute [tex]\( t = 4 \)[/tex] into [tex]\( h(t) = -16t^2 + 128t \)[/tex]:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
2. Calculate:
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 = 256 \][/tex]
The maximum height of the rocket is 256 feet.
I hope this helps you understand how to approach these types of problems! Let me know if you have any questions.
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