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Answer :
To find the maximum height of the projectile, we can analyze the path of the projectile represented by the equation:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This is a quadratic equation of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
In a quadratic equation like this, the projectile reaches its maximum height at the vertex of the parabola. For a parabola given by [tex]\( at^2 + bt + c \)[/tex], the time [tex]\( t \)[/tex] at which the maximum height occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} \][/tex]
[tex]\[ t = -\frac{48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
So, the projectile reaches its maximum height at [tex]\( t = 1.5 \)[/tex] seconds.
Now, to find the maximum height, we substitute [tex]\( t = 1.5 \)[/tex] back into the equation for [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculating each term:
1. [tex]\( -16(1.5)^2 = -16 \times 2.25 = -36 \)[/tex]
2. [tex]\( 48 \times 1.5 = 72 \)[/tex]
3. Constant term: [tex]\( 190 \)[/tex]
Add these together:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Therefore, the maximum height of the projectile is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This is a quadratic equation of the form [tex]\( h(t) = at^2 + bt + c \)[/tex], where:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
In a quadratic equation like this, the projectile reaches its maximum height at the vertex of the parabola. For a parabola given by [tex]\( at^2 + bt + c \)[/tex], the time [tex]\( t \)[/tex] at which the maximum height occurs is given by the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} \][/tex]
[tex]\[ t = -\frac{48}{-32} \][/tex]
[tex]\[ t = 1.5 \][/tex]
So, the projectile reaches its maximum height at [tex]\( t = 1.5 \)[/tex] seconds.
Now, to find the maximum height, we substitute [tex]\( t = 1.5 \)[/tex] back into the equation for [tex]\( h(t) \)[/tex]:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
Calculating each term:
1. [tex]\( -16(1.5)^2 = -16 \times 2.25 = -36 \)[/tex]
2. [tex]\( 48 \times 1.5 = 72 \)[/tex]
3. Constant term: [tex]\( 190 \)[/tex]
Add these together:
[tex]\[ h(1.5) = -36 + 72 + 190 = 226 \][/tex]
Therefore, the maximum height of the projectile is 226 feet.
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