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Answer :
To find the maximum height of a projectile, we need to analyze the provided quadratic equation that models its height over time:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This is a quadratic equation in the form of [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
The graph of this quadratic equation is a parabola, and 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 represents its maximum point.
The time at which the maximum height is reached can be found using the vertex formula for the time [tex]\( t \)[/tex] at the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the given coefficients [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \][/tex]
So, the projectile reaches its maximum height at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we substitute this value of [tex]\( t \)[/tex] back into the original height equation to find the maximum height:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Now, calculate:
1. [tex]\( (1.5)^2 = 2.25 \)[/tex]
2. [tex]\( -16 \times 2.25 = -36 \)[/tex]
3. [tex]\( 48 \times 1.5 = 72 \)[/tex]
Adding these together:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Thus, 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 of [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
The graph of this quadratic equation is a parabola, and 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 represents its maximum point.
The time at which the maximum height is reached can be found using the vertex formula for the time [tex]\( t \)[/tex] at the vertex of a parabola:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the given coefficients [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \][/tex]
So, the projectile reaches its maximum height at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we substitute this value of [tex]\( t \)[/tex] back into the original height equation to find the maximum height:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Now, calculate:
1. [tex]\( (1.5)^2 = 2.25 \)[/tex]
2. [tex]\( -16 \times 2.25 = -36 \)[/tex]
3. [tex]\( 48 \times 1.5 = 72 \)[/tex]
Adding these together:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Thus, the maximum height of the projectile is 226 feet.
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