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Answer :
To find out when the rocket hits the ground, we need to determine when the height of the rocket, [tex]\( h(t) \)[/tex], equals zero. The height function given is:
[tex]\[ h(t) = -16t^2 + 24t + 112 \][/tex]
This is a quadratic equation. The rocket hits the ground when its height is zero:
[tex]\[ -16t^2 + 24t + 112 = 0 \][/tex]
To find the value of [tex]\( t \)[/tex] when the rocket hits the ground, we solve this equation for [tex]\( t \)[/tex].
1. Write down the quadratic equation:
[tex]\(-16t^2 + 24t + 112 = 0\)[/tex]
2. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In the equation [tex]\( a = -16 \)[/tex], [tex]\( b = 24 \)[/tex], and [tex]\( c = 112 \)[/tex].
3. Calculate the discriminant:
[tex]\[ b^2 - 4ac = 24^2 - 4(-16)(112) \][/tex]
[tex]\[ b^2 - 4ac = 576 + 7168 = 7744 \][/tex]
4. Calculate the square root of the discriminant:
[tex]\(\sqrt{7744} = 88\)[/tex]
5. Plug in the values into the quadratic formula:
[tex]\[ t = \frac{-24 \pm 88}{-32} \][/tex]
We will get two potential solutions for [tex]\( t \)[/tex]:
- First solution: [tex]\( t = \frac{-24 + 88}{-32} = \frac{64}{-32} = -2 \)[/tex]
- Second solution: [tex]\( t = \frac{-24 - 88}{-32} = \frac{-112}{-32} = \frac{7}{2} = 3.5 \)[/tex]
Because time cannot be negative, the relevant solution is [tex]\( t = 3.5 \)[/tex].
Thus, the rocket hits the ground 3.5 seconds after launch.
[tex]\[ h(t) = -16t^2 + 24t + 112 \][/tex]
This is a quadratic equation. The rocket hits the ground when its height is zero:
[tex]\[ -16t^2 + 24t + 112 = 0 \][/tex]
To find the value of [tex]\( t \)[/tex] when the rocket hits the ground, we solve this equation for [tex]\( t \)[/tex].
1. Write down the quadratic equation:
[tex]\(-16t^2 + 24t + 112 = 0\)[/tex]
2. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In the equation [tex]\( a = -16 \)[/tex], [tex]\( b = 24 \)[/tex], and [tex]\( c = 112 \)[/tex].
3. Calculate the discriminant:
[tex]\[ b^2 - 4ac = 24^2 - 4(-16)(112) \][/tex]
[tex]\[ b^2 - 4ac = 576 + 7168 = 7744 \][/tex]
4. Calculate the square root of the discriminant:
[tex]\(\sqrt{7744} = 88\)[/tex]
5. Plug in the values into the quadratic formula:
[tex]\[ t = \frac{-24 \pm 88}{-32} \][/tex]
We will get two potential solutions for [tex]\( t \)[/tex]:
- First solution: [tex]\( t = \frac{-24 + 88}{-32} = \frac{64}{-32} = -2 \)[/tex]
- Second solution: [tex]\( t = \frac{-24 - 88}{-32} = \frac{-112}{-32} = \frac{7}{2} = 3.5 \)[/tex]
Because time cannot be negative, the relevant solution is [tex]\( t = 3.5 \)[/tex].
Thus, the rocket hits the ground 3.5 seconds after launch.
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