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Answer :
To find the maximum height of the projectile, we are dealing with a quadratic equation that describes its path:
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation is in the standard quadratic form:
[tex]\[ h(t) = at^2 + bt + c \][/tex]
where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
For a quadratic equation representing a parabola that opens downwards (which is the case when the coefficient of [tex]\( t^2 \)[/tex] is negative), the maximum value occurs at the vertex of the parabola. The time at which this maximum height occurs can be calculated using the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the formula, we get:
[tex]\[ t = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \][/tex]
So, the maximum height is reached at [tex]\( t = 1.5 \)[/tex] seconds.
Now, we substitute [tex]\( t = 1.5 \)[/tex] back into the original 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]
Then substitute back into the equation:
[tex]\[ h(1.5) = -16(2.25) + 48(1.5) + 190 \][/tex]
Calculate each term:
[tex]\[ -16(2.25) = -36 \][/tex]
[tex]\[ 48(1.5) = 72 \][/tex]
So the calculation becomes:
[tex]\[ h(1.5) = -36 + 72 + 190 \][/tex]
Adding these together gives:
[tex]\[ h(1.5) = 226 \][/tex]
Therefore, the maximum height of the projectile is 226 feet. So, the correct answer is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190 \][/tex]
This equation is in the standard quadratic form:
[tex]\[ h(t) = at^2 + bt + c \][/tex]
where [tex]\( a = -16 \)[/tex], [tex]\( b = 48 \)[/tex], and [tex]\( c = 190 \)[/tex].
For a quadratic equation representing a parabola that opens downwards (which is the case when the coefficient of [tex]\( t^2 \)[/tex] is negative), the maximum value occurs at the vertex of the parabola. The time at which this maximum height occurs can be calculated using the formula:
[tex]\[ t = -\frac{b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] into the formula, we get:
[tex]\[ t = -\frac{48}{2(-16)} = -\frac{48}{-32} = 1.5 \][/tex]
So, the maximum height is reached at [tex]\( t = 1.5 \)[/tex] seconds.
Now, we substitute [tex]\( t = 1.5 \)[/tex] back into the original 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]
Then substitute back into the equation:
[tex]\[ h(1.5) = -16(2.25) + 48(1.5) + 190 \][/tex]
Calculate each term:
[tex]\[ -16(2.25) = -36 \][/tex]
[tex]\[ 48(1.5) = 72 \][/tex]
So the calculation becomes:
[tex]\[ h(1.5) = -36 + 72 + 190 \][/tex]
Adding these together gives:
[tex]\[ h(1.5) = 226 \][/tex]
Therefore, the maximum height of the projectile is 226 feet. So, the correct answer is 226 feet.
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