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Answer :
To find the maximum height of a projectile, we can use the quadratic equation given for the projectile's path: [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex]. This is in the standard form of a quadratic equation [tex]\( at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
The maximum height occurs at the vertex of the parabola represented by this equation. For a quadratic equation, the time [tex]\( t \)[/tex] at which the maximum height is reached can be found using the formula for the vertex:
[tex]\[
t = -\frac{b}{2a}
\][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex]. Plug these values into the formula:
[tex]\[
t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5
\][/tex]
Now, to find the maximum height, substitute [tex]\( t = 1.5 \)[/tex] back into the original height equation:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Calculate each part step-by-step:
1. [tex]\( (1.5)^2 = 2.25 \)[/tex]
2. [tex]\( -16 \times 2.25 = -36 \)[/tex]
3. [tex]\( 48 \times 1.5 = 72 \)[/tex]
4. Then, add these results to the constant [tex]\( 190 \)[/tex]:
[tex]\[
h(1.5) = -36 + 72 + 190 = 226
\][/tex]
Thus, the maximum height of the projectile is 226 feet. Therefore, the correct answer is 226 feet.
The maximum height occurs at the vertex of the parabola represented by this equation. For a quadratic equation, the time [tex]\( t \)[/tex] at which the maximum height is reached can be found using the formula for the vertex:
[tex]\[
t = -\frac{b}{2a}
\][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex]. Plug these values into the formula:
[tex]\[
t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5
\][/tex]
Now, to find the maximum height, substitute [tex]\( t = 1.5 \)[/tex] back into the original height equation:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Calculate each part step-by-step:
1. [tex]\( (1.5)^2 = 2.25 \)[/tex]
2. [tex]\( -16 \times 2.25 = -36 \)[/tex]
3. [tex]\( 48 \times 1.5 = 72 \)[/tex]
4. Then, add these results to the constant [tex]\( 190 \)[/tex]:
[tex]\[
h(1.5) = -36 + 72 + 190 = 226
\][/tex]
Thus, the maximum height of the projectile is 226 feet. Therefore, the correct answer is 226 feet.
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