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Answer :
To find the maximum height of the projectile, we need to analyze the given equation for the projectile's path:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation is a quadratic function, which can be represented as a parabola. The maximum height of the projectile will occur at the vertex of this parabola. In a quadratic equation of the form [tex]\( ax^2 + bx + c \)[/tex], the t-coordinate of the vertex can be found using the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For this problem, the coefficients are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Let's calculate the time when the projectile reaches its maximum height:
1. Identify the coefficients: [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex].
2. Use the formula for the time at the vertex:
[tex]\[ t = -\frac{b}{2a} = -\frac{48}{2 \times (-16)} = \frac{48}{32} = 1.5 \][/tex]
Now that we know the time, [tex]\( t = 1.5 \)[/tex] seconds, when the maximum height occurs, we substitute it back into the height equation to find the maximum height:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Let's compute the values step-by-step:
1. Calculate [tex]\( (1.5)^2 = 2.25 \)[/tex].
2. Find the first term: [tex]\(-16 \times 2.25 = -36\)[/tex].
3. Find the second term: [tex]\(48 \times 1.5 = 72\)[/tex].
Substitute these back into the equation:
[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 is a quadratic function, which can be represented as a parabola. The maximum height of the projectile will occur at the vertex of this parabola. In a quadratic equation of the form [tex]\( ax^2 + bx + c \)[/tex], the t-coordinate of the vertex can be found using the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For this problem, the coefficients are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Let's calculate the time when the projectile reaches its maximum height:
1. Identify the coefficients: [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex].
2. Use the formula for the time at the vertex:
[tex]\[ t = -\frac{b}{2a} = -\frac{48}{2 \times (-16)} = \frac{48}{32} = 1.5 \][/tex]
Now that we know the time, [tex]\( t = 1.5 \)[/tex] seconds, when the maximum height occurs, we substitute it back into the height equation to find the maximum height:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Let's compute the values step-by-step:
1. Calculate [tex]\( (1.5)^2 = 2.25 \)[/tex].
2. Find the first term: [tex]\(-16 \times 2.25 = -36\)[/tex].
3. Find the second term: [tex]\(48 \times 1.5 = 72\)[/tex].
Substitute these back into the equation:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Thus, the maximum height of the projectile is 226 feet.
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