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Answer :
Sure! Let's solve the problem step by step:
We have the function for the height of a toy rocket given by [tex]\( h(t) = -16t^2 + 128t \)[/tex], where [tex]\( t \)[/tex] is the time in seconds, and [tex]\( h(t) \)[/tex] is the height in feet.
### a. Time to reach maximum height:
To find the time at which the rocket reaches its maximum height, we need to find the vertex of the parabola described by the quadratic equation. The formula to find the time at which the maximum height is reached for a quadratic [tex]\( at^2 + bt + c \)[/tex] is:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex].
[tex]\[ t = -\frac{128}{2 \times -16} = 4 \text{ seconds} \][/tex]
### b. Maximum height:
To find the maximum height, we substitute the value of [tex]\( t \)[/tex] from above into the height equation [tex]\( h(t) \)[/tex].
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \text{ feet} \][/tex]
### c. Time to reach the ground:
The rocket reaches the ground when its height is 0, so we need to solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor the equation:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( t = 0 \)[/tex] (the time at launch)
2. [tex]\( -16t + 128 = 0 \)[/tex]
Solving [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ 16t = 128 \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \text{ seconds} \][/tex]
So the rocket reaches the ground after 8 seconds.
In summary:
- The rocket takes 4 seconds to reach its maximum height.
- The maximum height is 256 feet.
- The rocket reaches the ground after 8 seconds.
We have the function for the height of a toy rocket given by [tex]\( h(t) = -16t^2 + 128t \)[/tex], where [tex]\( t \)[/tex] is the time in seconds, and [tex]\( h(t) \)[/tex] is the height in feet.
### a. Time to reach maximum height:
To find the time at which the rocket reaches its maximum height, we need to find the vertex of the parabola described by the quadratic equation. The formula to find the time at which the maximum height is reached for a quadratic [tex]\( at^2 + bt + c \)[/tex] is:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex].
[tex]\[ t = -\frac{128}{2 \times -16} = 4 \text{ seconds} \][/tex]
### b. Maximum height:
To find the maximum height, we substitute the value of [tex]\( t \)[/tex] from above into the height equation [tex]\( h(t) \)[/tex].
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
[tex]\[ h(4) = -16 \times 16 + 128 \times 4 \][/tex]
[tex]\[ h(4) = -256 + 512 \][/tex]
[tex]\[ h(4) = 256 \text{ feet} \][/tex]
### c. Time to reach the ground:
The rocket reaches the ground when its height is 0, so we need to solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor the equation:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( t = 0 \)[/tex] (the time at launch)
2. [tex]\( -16t + 128 = 0 \)[/tex]
Solving [tex]\( -16t + 128 = 0 \)[/tex]:
[tex]\[ 16t = 128 \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \text{ seconds} \][/tex]
So the rocket reaches the ground after 8 seconds.
In summary:
- The rocket takes 4 seconds to reach its maximum height.
- The maximum height is 256 feet.
- The rocket reaches the ground after 8 seconds.
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