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Answer :
To find the maximum height of the projectile, we'll need to look at the equation given for its height over time:
[tex]\[ h(t) = -16t^2 + 48t + 190. \][/tex]
This is a quadratic equation, and its graph is a parabola that opens downwards (since the coefficient of [tex]\( t^2 \)[/tex] is negative). The maximum height of a parabola that opens downwards occurs at its vertex.
The formula to find the time at which the maximum height occurs for a quadratic equation [tex]\( at^2 + bt + c \)[/tex] is:
[tex]\[ t = -\frac{b}{2a}, \][/tex]
where [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex] in our equation. Let's calculate this:
[tex]\[ t = -\frac{48}{2 \times (-16)} = -\frac{48}{-32} = \frac{48}{32} = 1.5 \text{ seconds}. \][/tex]
Next, we substitute [tex]\( t = 1.5 \)[/tex] back into the original equation to find the height at this time:
[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 back:
[tex]\[ h(1.5) = -16(2.25) + 48(1.5) + 190. \][/tex]
Calculate each term:
1. [tex]\( -16(2.25) = -36 \)[/tex].
2. [tex]\( 48(1.5) = 72 \)[/tex].
Now add them up:
[tex]\[ h(1.5) = -36 + 72 + 190. \][/tex]
[tex]\[ h(1.5) = 226. \][/tex]
So, the maximum height of the projectile is 226 feet.
[tex]\[ h(t) = -16t^2 + 48t + 190. \][/tex]
This is a quadratic equation, and its graph is a parabola that opens downwards (since the coefficient of [tex]\( t^2 \)[/tex] is negative). The maximum height of a parabola that opens downwards occurs at its vertex.
The formula to find the time at which the maximum height occurs for a quadratic equation [tex]\( at^2 + bt + c \)[/tex] is:
[tex]\[ t = -\frac{b}{2a}, \][/tex]
where [tex]\( a = -16 \)[/tex] and [tex]\( b = 48 \)[/tex] in our equation. Let's calculate this:
[tex]\[ t = -\frac{48}{2 \times (-16)} = -\frac{48}{-32} = \frac{48}{32} = 1.5 \text{ seconds}. \][/tex]
Next, we substitute [tex]\( t = 1.5 \)[/tex] back into the original equation to find the height at this time:
[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 back:
[tex]\[ h(1.5) = -16(2.25) + 48(1.5) + 190. \][/tex]
Calculate each term:
1. [tex]\( -16(2.25) = -36 \)[/tex].
2. [tex]\( 48(1.5) = 72 \)[/tex].
Now add them up:
[tex]\[ h(1.5) = -36 + 72 + 190. \][/tex]
[tex]\[ h(1.5) = 226. \][/tex]
So, the maximum height of the projectile is 226 feet.
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