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Answer :
To find the maximum height of the projectile, we need to analyze the path modeled by the equation [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex]. This equation represents a parabola, and because the coefficient of [tex]\( t^2 \)[/tex] is negative, the parabola opens downward. The maximum height will therefore occur at the vertex of the parabola.
Here is a step-by-step guide to finding the maximum height:
1. Identify the Form of the Equation:
The quadratic equation is in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
2. Calculate the Time at Vertex:
The formula to find the time [tex]\( t \)[/tex] at which the maximum height occurs (the vertex) for a quadratic in the form [tex]\( at^2 + bt + c \)[/tex] is given by:
[tex]\[
t = -\frac{b}{2a}
\][/tex]
Plug in the values of [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[
t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \text{ seconds}
\][/tex]
3. Determine the Maximum Height:
Substitute [tex]\( t = 1.5 \)[/tex] back into the original equation to find the maximum height:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Calculate [tex]\( (1.5)^2 \)[/tex]:
[tex]\[
(1.5)^2 = 2.25
\][/tex]
Now substitute:
[tex]\[
h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190
\][/tex]
Compute each term:
[tex]\[
-16 \times 2.25 = -36
\][/tex]
[tex]\[
48 \times 1.5 = 72
\][/tex]
Finally, add these results together:
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
Therefore, the maximum height of the projectile is 226 feet.
Here is a step-by-step guide to finding the maximum height:
1. Identify the Form of the Equation:
The quadratic equation is in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
2. Calculate the Time at Vertex:
The formula to find the time [tex]\( t \)[/tex] at which the maximum height occurs (the vertex) for a quadratic in the form [tex]\( at^2 + bt + c \)[/tex] is given by:
[tex]\[
t = -\frac{b}{2a}
\][/tex]
Plug in the values of [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[
t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \text{ seconds}
\][/tex]
3. Determine the Maximum Height:
Substitute [tex]\( t = 1.5 \)[/tex] back into the original equation to find the maximum height:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Calculate [tex]\( (1.5)^2 \)[/tex]:
[tex]\[
(1.5)^2 = 2.25
\][/tex]
Now substitute:
[tex]\[
h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190
\][/tex]
Compute each term:
[tex]\[
-16 \times 2.25 = -36
\][/tex]
[tex]\[
48 \times 1.5 = 72
\][/tex]
Finally, add these results together:
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
Therefore, the maximum height of the projectile is 226 feet.
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