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Answer :
To determine the maximum height of the projectile, we can use the path equation of the projectile:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation represents a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
In a quadratic equation, the maximum or minimum value (vertex) can be found using the formula for the time at which the vertex occurs:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Now, let's calculate the time at which the maximum height occurs:
1. Identify the coefficients:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
2. Plug these values into the formula to find [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \][/tex]
So, the maximum height occurs at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we find the maximum height by plugging [tex]\( t = 1.5 \)[/tex] back into the height equation [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculate each term:
- [tex]\( (1.5)^2 = 2.25 \)[/tex]
- Multiply by [tex]\(-16\)[/tex]: [tex]\(-16 \times 2.25 = -36\)[/tex]
- Multiply [tex]\( 48 \times 1.5 = 72 \)[/tex]
Now add them:
[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 equation represents a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
In a quadratic equation, the maximum or minimum value (vertex) can be found using the formula for the time at which the vertex occurs:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Now, let's calculate the time at which the maximum height occurs:
1. Identify the coefficients:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
2. Plug these values into the formula to find [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \][/tex]
So, the maximum height occurs at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we find the maximum height by plugging [tex]\( t = 1.5 \)[/tex] back into the height equation [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculate each term:
- [tex]\( (1.5)^2 = 2.25 \)[/tex]
- Multiply by [tex]\(-16\)[/tex]: [tex]\(-16 \times 2.25 = -36\)[/tex]
- Multiply [tex]\( 48 \times 1.5 = 72 \)[/tex]
Now add them:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Thus, the maximum height of the projectile is 226 feet.
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