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Answer :
Sure! Let's solve the problem step-by-step.
We have the height of a toy rocket given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
where [tex]\( t \)[/tex] is the time in seconds.
### Part a: Time to reach maximum height
The height function is a parabola that opens downwards (since the coefficient of [tex]\( t^2 \)[/tex] is negative). The maximum height is at the vertex. The time [tex]\( t \)[/tex] at which the maximum height occurs is given by the formula for the vertex of a quadratic equation:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, the coefficients are [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex]. Plugging in these values:
[tex]\[ t = -\frac{128}{2 \times (-16)} = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### Part b: Maximum height
To find the maximum height, substitute [tex]\( t = 4 \)[/tex] back into the height equation:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
Calculate each term:
- [tex]\( 4^2 = 16 \)[/tex]
- [tex]\( -16 \times 16 = -256 \)[/tex]
- [tex]\( 128 \times 4 = 512 \)[/tex]
Add them together:
[tex]\[ h(4) = -256 + 512 = 256 \][/tex]
Thus, the maximum height is 256 feet.
### Part c: Time to reach the ground
The rocket reaches the ground when its height [tex]\( h(t) \)[/tex] is 0. We solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives two solutions:
1. [tex]\( t = 0 \)[/tex]: The time at launch.
2. [tex]\( -16t + 128 = 0 \)[/tex]: Solve for [tex]\( t \)[/tex].
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, it takes 8 seconds for the rocket to reach the ground.
In summary:
- The rocket reaches its maximum height at 4 seconds.
- The maximum height is 256 feet.
- The rocket takes 8 seconds to come back to the ground.
We have the height of a toy rocket given by the equation:
[tex]\[ h(t) = -16t^2 + 128t \][/tex]
where [tex]\( t \)[/tex] is the time in seconds.
### Part a: Time to reach maximum height
The height function is a parabola that opens downwards (since the coefficient of [tex]\( t^2 \)[/tex] is negative). The maximum height is at the vertex. The time [tex]\( t \)[/tex] at which the maximum height occurs is given by the formula for the vertex of a quadratic equation:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Here, the coefficients are [tex]\( a = -16 \)[/tex] and [tex]\( b = 128 \)[/tex]. Plugging in these values:
[tex]\[ t = -\frac{128}{2 \times (-16)} = 4 \][/tex]
So, it takes 4 seconds for the rocket to reach its maximum height.
### Part b: Maximum height
To find the maximum height, substitute [tex]\( t = 4 \)[/tex] back into the height equation:
[tex]\[ h(4) = -16(4)^2 + 128(4) \][/tex]
Calculate each term:
- [tex]\( 4^2 = 16 \)[/tex]
- [tex]\( -16 \times 16 = -256 \)[/tex]
- [tex]\( 128 \times 4 = 512 \)[/tex]
Add them together:
[tex]\[ h(4) = -256 + 512 = 256 \][/tex]
Thus, the maximum height is 256 feet.
### Part c: Time to reach the ground
The rocket reaches the ground when its height [tex]\( h(t) \)[/tex] is 0. We solve the equation:
[tex]\[ -16t^2 + 128t = 0 \][/tex]
Factor out [tex]\( t \)[/tex]:
[tex]\[ t(-16t + 128) = 0 \][/tex]
This gives two solutions:
1. [tex]\( t = 0 \)[/tex]: The time at launch.
2. [tex]\( -16t + 128 = 0 \)[/tex]: Solve for [tex]\( t \)[/tex].
[tex]\[ -16t = -128 \][/tex]
[tex]\[ t = \frac{128}{16} = 8 \][/tex]
So, it takes 8 seconds for the rocket to reach the ground.
In summary:
- The rocket reaches its maximum height at 4 seconds.
- The maximum height is 256 feet.
- The rocket takes 8 seconds to come back to the ground.
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