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Answer :
To find the maximum height of the projectile, we need to examine the equation that models its height over time:
[tex]\[ h(t) = -16t^2 + 48t + 190. \][/tex]
This equation is a quadratic function, which represents a parabola. The maximum height of the projectile is reached at the vertex of this parabola since the parabola opens downward (indicated by the negative coefficient of [tex]\( t^2 \)[/tex]).
For a quadratic equation of the form [tex]\( ax^2 + bx + c \)[/tex], the time at which the vertex (and hence the maximum point for a downward-opening parabola) occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a}. \][/tex]
In this equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Substituting these values into the formula gives:
[tex]\[ t = -\frac{48}{2 \times -16}. \][/tex]
First, calculate the denominator:
[tex]\[ 2 \times -16 = -32. \][/tex]
Now, calculate the value of [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{-32} = \frac{48}{32} = 1.5. \][/tex]
The time at which the maximum height occurs is 1.5 seconds. Now, we substitute [tex]\( t = 1.5 \)[/tex] into the height 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 this back into the equation:
[tex]\[ h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190. \][/tex]
Perform the multiplications:
[tex]\[ -16 \times 2.25 = -36, \][/tex]
[tex]\[ 48 \times 1.5 = 72. \][/tex]
Add these results along with 190:
[tex]\[ h(1.5) = -36 + 72 + 190. \][/tex]
Now, perform the additions:
[tex]\[ h(1.5) = 36 + 190 = 226. \][/tex]
Thus, the maximum height of the projectile is 226 feet. The correct answer is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190. \][/tex]
This equation is a quadratic function, which represents a parabola. The maximum height of the projectile is reached at the vertex of this parabola since the parabola opens downward (indicated by the negative coefficient of [tex]\( t^2 \)[/tex]).
For a quadratic equation of the form [tex]\( ax^2 + bx + c \)[/tex], the time at which the vertex (and hence the maximum point for a downward-opening parabola) occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a}. \][/tex]
In this equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Substituting these values into the formula gives:
[tex]\[ t = -\frac{48}{2 \times -16}. \][/tex]
First, calculate the denominator:
[tex]\[ 2 \times -16 = -32. \][/tex]
Now, calculate the value of [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{-32} = \frac{48}{32} = 1.5. \][/tex]
The time at which the maximum height occurs is 1.5 seconds. Now, we substitute [tex]\( t = 1.5 \)[/tex] into the height 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 this back into the equation:
[tex]\[ h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190. \][/tex]
Perform the multiplications:
[tex]\[ -16 \times 2.25 = -36, \][/tex]
[tex]\[ 48 \times 1.5 = 72. \][/tex]
Add these results along with 190:
[tex]\[ h(1.5) = -36 + 72 + 190. \][/tex]
Now, perform the additions:
[tex]\[ h(1.5) = 36 + 190 = 226. \][/tex]
Thus, the maximum height of the projectile is 226 feet. The correct answer is 226 feet.
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