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Answer :
To determine how long it will take the toy rocket to reach the ground, we need to set the height equation to zero because the rocket is on the ground when its height is zero. The height equation given is:
[tex]\[ h(t) = -16t^2 + 128t + 12 \][/tex]
We need to find the time [tex]\( t \)[/tex] when [tex]\( h(t) = 0 \)[/tex].
So, set up the equation:
[tex]\[ -16t^2 + 128t + 12 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 128 \)[/tex], and [tex]\( c = 12 \)[/tex].
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plug the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4(-16)(12)}}{2(-16)} \][/tex]
Calculate the discriminant [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ 128^2 = 16384 \][/tex]
[tex]\[ 4(-16)(12) = -768 \][/tex]
[tex]\[ 16384 + 768 = 17152 \][/tex]
Now, calculate the square root of the discriminant:
[tex]\[ \sqrt{17152} \approx 130.923 \][/tex]
Substitute back into the quadratic formula:
[tex]\[ t = \frac{-128 \pm 130.923}{-32} \][/tex]
This provides two solutions:
1. [tex]\( t = \frac{-128 + 130.923}{-32} \)[/tex]
2. [tex]\( t = \frac{-128 - 130.923}{-32} \)[/tex]
Calculate these solutions:
1. [tex]\( t \approx \frac{2.923}{-32} \approx -0.091 \)[/tex]
2. [tex]\( t \approx \frac{-258.923}{-32} \approx 8.0927 \)[/tex]
Since time cannot be negative, we take the positive solution:
Hence, the rocket will reach the ground in approximately 8.1 seconds, rounded to the nearest tenth of a second.
[tex]\[ h(t) = -16t^2 + 128t + 12 \][/tex]
We need to find the time [tex]\( t \)[/tex] when [tex]\( h(t) = 0 \)[/tex].
So, set up the equation:
[tex]\[ -16t^2 + 128t + 12 = 0 \][/tex]
This is a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 128 \)[/tex], and [tex]\( c = 12 \)[/tex].
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plug the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4(-16)(12)}}{2(-16)} \][/tex]
Calculate the discriminant [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ 128^2 = 16384 \][/tex]
[tex]\[ 4(-16)(12) = -768 \][/tex]
[tex]\[ 16384 + 768 = 17152 \][/tex]
Now, calculate the square root of the discriminant:
[tex]\[ \sqrt{17152} \approx 130.923 \][/tex]
Substitute back into the quadratic formula:
[tex]\[ t = \frac{-128 \pm 130.923}{-32} \][/tex]
This provides two solutions:
1. [tex]\( t = \frac{-128 + 130.923}{-32} \)[/tex]
2. [tex]\( t = \frac{-128 - 130.923}{-32} \)[/tex]
Calculate these solutions:
1. [tex]\( t \approx \frac{2.923}{-32} \approx -0.091 \)[/tex]
2. [tex]\( t \approx \frac{-258.923}{-32} \approx 8.0927 \)[/tex]
Since time cannot be negative, we take the positive solution:
Hence, the rocket will reach the ground in approximately 8.1 seconds, rounded to the nearest tenth of a second.
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