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Answer :
To find out how long it takes for the rocket to reach the ground, we need to determine when the height [tex]\( h(t) \)[/tex] of the rocket is 0. The height of the rocket at any time [tex]\( t \)[/tex] seconds after launch is given by the equation:
[tex]\[ h(t) = -16t^2 + 128t + 12 \][/tex]
We set [tex]\( h(t) = 0 \)[/tex] to find when the rocket reaches the ground:
[tex]\[ 0 = -16t^2 + 128t + 12 \][/tex]
This is a quadratic equation in 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]
Plugging in the values, we have:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4 \times (-16) \times 12}}{2 \times (-16)} \][/tex]
1. Calculate the discriminant [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ 128^2 = 16384 \][/tex]
[tex]\[ 4 \times (-16) \times 12 = -768 \][/tex]
[tex]\[ b^2 - 4ac = 16384 + 768 = 17152 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{17152} = \text{approximately } 131 \][/tex]
3. Substitute back to find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-128 \pm 131}{-32} \][/tex]
This gives us two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-128 + 131}{-32} = \frac{3}{-32} = \text{approximately } 4 - \frac{\sqrt{67}}{2} \][/tex]
[tex]\[ t_2 = \frac{-128 - 131}{-32} = -\frac{259}{-32} = \text{approximately } 4 + \frac{\sqrt{67}}{2} \][/tex]
Since time cannot be negative, the valid solution for when the rocket reaches the ground is approximately [tex]\( t = 4 + \frac{\sqrt{67}}{2} \)[/tex].
[tex]\[ h(t) = -16t^2 + 128t + 12 \][/tex]
We set [tex]\( h(t) = 0 \)[/tex] to find when the rocket reaches the ground:
[tex]\[ 0 = -16t^2 + 128t + 12 \][/tex]
This is a quadratic equation in 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]
Plugging in the values, we have:
[tex]\[ t = \frac{-128 \pm \sqrt{128^2 - 4 \times (-16) \times 12}}{2 \times (-16)} \][/tex]
1. Calculate the discriminant [tex]\( b^2 - 4ac \)[/tex]:
[tex]\[ 128^2 = 16384 \][/tex]
[tex]\[ 4 \times (-16) \times 12 = -768 \][/tex]
[tex]\[ b^2 - 4ac = 16384 + 768 = 17152 \][/tex]
2. Find the square root of the discriminant:
[tex]\[ \sqrt{17152} = \text{approximately } 131 \][/tex]
3. Substitute back to find [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-128 \pm 131}{-32} \][/tex]
This gives us two potential solutions for [tex]\( t \)[/tex]:
[tex]\[ t_1 = \frac{-128 + 131}{-32} = \frac{3}{-32} = \text{approximately } 4 - \frac{\sqrt{67}}{2} \][/tex]
[tex]\[ t_2 = \frac{-128 - 131}{-32} = -\frac{259}{-32} = \text{approximately } 4 + \frac{\sqrt{67}}{2} \][/tex]
Since time cannot be negative, the valid solution for when the rocket reaches the ground is approximately [tex]\( t = 4 + \frac{\sqrt{67}}{2} \)[/tex].
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