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Answer :
To find the maximum height of the projectile, we need to determine the highest point on the path described by the quadratic equation [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex].
This is a quadratic equation in the form of [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
For a quadratic equation, the maximum or minimum value occurs at the vertex, which is at time [tex]\( t = -\frac{b}{2a} \)[/tex].
1. Calculate the time at which the maximum height occurs:
[tex]\[
t = -\frac{b}{2a} = -\frac{48}{2 \times (-16)} = -\frac{48}{-32} = 1.5 \, \text{seconds}
\][/tex]
2. Calculate the maximum height:
Substitute [tex]\( t = 1.5 \)[/tex] into the height equation:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
[tex]\[
h(1.5) = -16(2.25) + 72 + 190
\][/tex]
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
The maximum height reached by the projectile is 226 feet. Therefore, the correct answer is 226 feet.
This is a quadratic equation in the form of [tex]\( ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
For a quadratic equation, the maximum or minimum value occurs at the vertex, which is at time [tex]\( t = -\frac{b}{2a} \)[/tex].
1. Calculate the time at which the maximum height occurs:
[tex]\[
t = -\frac{b}{2a} = -\frac{48}{2 \times (-16)} = -\frac{48}{-32} = 1.5 \, \text{seconds}
\][/tex]
2. Calculate the maximum height:
Substitute [tex]\( t = 1.5 \)[/tex] into the height equation:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
[tex]\[
h(1.5) = -16(2.25) + 72 + 190
\][/tex]
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
The maximum height reached by the projectile is 226 feet. Therefore, the correct answer is 226 feet.
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