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Answer :
To find the maximum height of the projectile, we need to analyze the given quadratic equation that models its path:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation represents the height [tex]\( h \)[/tex] of the projectile at time [tex]\( t \)[/tex]. The coefficients of the quadratic equation are related to the path of the projectile:
1. The coefficient of [tex]\( t^2 \)[/tex], which is [tex]\(-16\)[/tex], indicates the effect of gravity. This negative sign shows the parabola opens downward.
2. The coefficient of [tex]\( t \)[/tex], which is [tex]\(48\)[/tex], represents the initial velocity of the projectile.
3. The constant term, [tex]\(190\)[/tex], is the initial height from which it was launched.
To find the maximum height, we need to determine the vertex of the parabola represented by this equation. The vertex will give us the maximum point on the parabola because this is a downward-opening parabola.
For a quadratic equation in the form [tex]\( at^2 + bt + c \)[/tex], the time [tex]\( t \)[/tex] at which the maximum height occurs is given by:
[tex]\[ t = \frac{-b}{2a} \][/tex]
Plugging the values from the equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
The time is:
[tex]\[ t = \frac{-48}{2 \times (-16)} \][/tex]
[tex]\[ t = \frac{-48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
At [tex]\( t = 1.5 \)[/tex] seconds, the projectile reaches its maximum height. Now, to find the maximum height, substitute [tex]\( t = 1.5 \)[/tex] back into the height equation:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Simplifying, we find:
[tex]\[ h(1.5) = -16(2.25) + 72 + 190 \][/tex]
[tex]\[ h(1.5) = -36 + 72 + 190 \][/tex]
[tex]\[ h(1.5) = 226 \][/tex]
Thus, the maximum height of the projectile is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation represents the height [tex]\( h \)[/tex] of the projectile at time [tex]\( t \)[/tex]. The coefficients of the quadratic equation are related to the path of the projectile:
1. The coefficient of [tex]\( t^2 \)[/tex], which is [tex]\(-16\)[/tex], indicates the effect of gravity. This negative sign shows the parabola opens downward.
2. The coefficient of [tex]\( t \)[/tex], which is [tex]\(48\)[/tex], represents the initial velocity of the projectile.
3. The constant term, [tex]\(190\)[/tex], is the initial height from which it was launched.
To find the maximum height, we need to determine the vertex of the parabola represented by this equation. The vertex will give us the maximum point on the parabola because this is a downward-opening parabola.
For a quadratic equation in the form [tex]\( at^2 + bt + c \)[/tex], the time [tex]\( t \)[/tex] at which the maximum height occurs is given by:
[tex]\[ t = \frac{-b}{2a} \][/tex]
Plugging the values from the equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
The time is:
[tex]\[ t = \frac{-48}{2 \times (-16)} \][/tex]
[tex]\[ t = \frac{-48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
At [tex]\( t = 1.5 \)[/tex] seconds, the projectile reaches its maximum height. Now, to find the maximum height, substitute [tex]\( t = 1.5 \)[/tex] back into the height equation:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Simplifying, we find:
[tex]\[ h(1.5) = -16(2.25) + 72 + 190 \][/tex]
[tex]\[ h(1.5) = -36 + 72 + 190 \][/tex]
[tex]\[ h(1.5) = 226 \][/tex]
Thus, the maximum height of the projectile is 226 feet.
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