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Answer :
To find the maximum height of the projectile, we need to analyze the motion described by the equation of its path:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This is a quadratic equation in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
The equation of the path represents a parabola, and since [tex]\( a = -16 \)[/tex] (which is negative), the parabola opens downward. This means the maximum height is at the vertex of the parabola.
To find the time at which the maximum height occurs, we use the formula to find the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} \][/tex]
[tex]\[ t = -\frac{48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
Now, we 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]
First, calculate [tex]\( (1.5)^2 \)[/tex]:
[tex]\[ (1.5)^2 = 2.25 \][/tex]
Now, substitute and calculate the height:
[tex]\[ h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190 \][/tex]
[tex]\[ h(1.5) = -36 + 72 + 190 \][/tex]
[tex]\[ h(1.5) = 226 \][/tex]
Therefore, the maximum height of the projectile is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This is a quadratic equation in the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
The equation of the path represents a parabola, and since [tex]\( a = -16 \)[/tex] (which is negative), the parabola opens downward. This means the maximum height is at the vertex of the parabola.
To find the time at which the maximum height occurs, we use the formula to find the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} \][/tex]
[tex]\[ t = -\frac{48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
Now, we 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]
First, calculate [tex]\( (1.5)^2 \)[/tex]:
[tex]\[ (1.5)^2 = 2.25 \][/tex]
Now, substitute and calculate the height:
[tex]\[ h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190 \][/tex]
[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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