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Answer :
To find the maximum height of a projectile launched from a building, we can use the equation that models the height of the projectile over time. The equation is [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex].
1. Understanding the Equation:
- [tex]\( h(t) \)[/tex] represents the height of the projectile at time [tex]\( t \)[/tex].
- The equation has the form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
- Since the coefficient [tex]\( a \)[/tex] is negative, the parabola opens downwards, which means it has a maximum point.
2. Finding the Time of Maximum Height:
- The time at which the maximum height occurs can be found using the formula for the vertex of a parabola, [tex]\( t = -\frac{b}{2a} \)[/tex].
- Substitute the values: [tex]\( b = 48 \)[/tex] and [tex]\( a = -16 \)[/tex].
3. Calculating [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.
4. Finding the Maximum Height:
- Substitute [tex]\( t = 1.5 \)[/tex] back into the height equation to find [tex]\( h(1.5) \)[/tex].
- [tex]\( h(1.5) = -16 \times (1.5)^2 + 48 \times 1.5 + 190 \)[/tex].
5. Calculating the Maximum Height:
- Compute the expression:
- [tex]\( -16 \times (1.5)^2 = -16 \times 2.25 = -36 \)[/tex].
- [tex]\( 48 \times 1.5 = 72 \)[/tex].
- Add these to [tex]\( 190 \)[/tex]:
[tex]\( h(1.5) = -36 + 72 + 190 = 226 \)[/tex] feet.
Thus, the maximum height of the projectile is 226 feet.
1. Understanding the Equation:
- [tex]\( h(t) \)[/tex] represents the height of the projectile at time [tex]\( t \)[/tex].
- The equation has the form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
- Since the coefficient [tex]\( a \)[/tex] is negative, the parabola opens downwards, which means it has a maximum point.
2. Finding the Time of Maximum Height:
- The time at which the maximum height occurs can be found using the formula for the vertex of a parabola, [tex]\( t = -\frac{b}{2a} \)[/tex].
- Substitute the values: [tex]\( b = 48 \)[/tex] and [tex]\( a = -16 \)[/tex].
3. Calculating [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.
4. Finding the Maximum Height:
- Substitute [tex]\( t = 1.5 \)[/tex] back into the height equation to find [tex]\( h(1.5) \)[/tex].
- [tex]\( h(1.5) = -16 \times (1.5)^2 + 48 \times 1.5 + 190 \)[/tex].
5. Calculating the Maximum Height:
- Compute the expression:
- [tex]\( -16 \times (1.5)^2 = -16 \times 2.25 = -36 \)[/tex].
- [tex]\( 48 \times 1.5 = 72 \)[/tex].
- Add these to [tex]\( 190 \)[/tex]:
[tex]\( h(1.5) = -36 + 72 + 190 = 226 \)[/tex] feet.
Thus, the maximum height of the projectile is 226 feet.
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