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Answer :
To find the maximum height of the projectile, we need to analyze the given equation for the height of the projectile over time:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation is in the form of a quadratic equation, [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
The graph of a quadratic equation is a parabola. Since the coefficient of [tex]\( t^2 \)[/tex] (which is [tex]\( a = -16 \)[/tex]) is negative, the parabola opens downward. This means the vertex of the parabola will give us the maximum height of the projectile.
To find the vertex, we use the formula for the [tex]\( t \)[/tex]-coordinate of the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \][/tex]
At [tex]\( t = 1.5 \)[/tex] seconds, the projectile reaches its maximum height. Now, we plug [tex]\( t = 1.5 \)[/tex] back into the height equation to find the maximum height [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculating step-by-step:
- [tex]\((1.5)^2 = 2.25\)[/tex]
- [tex]\(-16 \times 2.25 = -36\)[/tex]
- [tex]\(48 \times 1.5 = 72\)[/tex]
- Add them all together: [tex]\(-36 + 72 + 190 = 226\)[/tex]
Thus, the maximum height of the projectile is 226 feet.
The correct answer from the choices is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation is in the form of a quadratic equation, [tex]\( h(t) = at^2 + bt + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
The graph of a quadratic equation is a parabola. Since the coefficient of [tex]\( t^2 \)[/tex] (which is [tex]\( a = -16 \)[/tex]) is negative, the parabola opens downward. This means the vertex of the parabola will give us the maximum height of the projectile.
To find the vertex, we use the formula for the [tex]\( t \)[/tex]-coordinate of the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( b \)[/tex] and [tex]\( a \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \][/tex]
At [tex]\( t = 1.5 \)[/tex] seconds, the projectile reaches its maximum height. Now, we plug [tex]\( t = 1.5 \)[/tex] back into the height equation to find the maximum height [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculating step-by-step:
- [tex]\((1.5)^2 = 2.25\)[/tex]
- [tex]\(-16 \times 2.25 = -36\)[/tex]
- [tex]\(48 \times 1.5 = 72\)[/tex]
- Add them all together: [tex]\(-36 + 72 + 190 = 226\)[/tex]
Thus, the maximum height of the projectile is 226 feet.
The correct answer from the choices is 226 feet.
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