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Answer :
The projectile's height is given by the quadratic equation
[tex]$$
h(t) = -16t^2 + 48t + 190.
$$[/tex]
Since the coefficient of [tex]$t^2$[/tex] is negative, the graph is a downward-opening parabola, and its maximum height occurs at the vertex.
The time [tex]$t$[/tex] at which the vertex occurs is found using the formula
[tex]$$
t = -\frac{b}{2a},
$$[/tex]
where for our equation [tex]$a = -16$[/tex] and [tex]$b = 48$[/tex]. Substituting these values, we get
[tex]$$
t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5 \text{ seconds}.
$$[/tex]
Now, to find the maximum height, substitute [tex]$t = 1.5$[/tex] back into the equation:
[tex]$$
\begin{aligned}
h(1.5) &= -16(1.5)^2 + 48(1.5) + 190 \\
&= -16(2.25) + 72 + 190 \\
&= -36 + 72 + 190 \\
&= 226 \text{ feet}.
\end{aligned}
$$[/tex]
Thus, the maximum height of the projectile is [tex]$\boxed{226 \text{ feet}}$[/tex].
[tex]$$
h(t) = -16t^2 + 48t + 190.
$$[/tex]
Since the coefficient of [tex]$t^2$[/tex] is negative, the graph is a downward-opening parabola, and its maximum height occurs at the vertex.
The time [tex]$t$[/tex] at which the vertex occurs is found using the formula
[tex]$$
t = -\frac{b}{2a},
$$[/tex]
where for our equation [tex]$a = -16$[/tex] and [tex]$b = 48$[/tex]. Substituting these values, we get
[tex]$$
t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5 \text{ seconds}.
$$[/tex]
Now, to find the maximum height, substitute [tex]$t = 1.5$[/tex] back into the equation:
[tex]$$
\begin{aligned}
h(1.5) &= -16(1.5)^2 + 48(1.5) + 190 \\
&= -16(2.25) + 72 + 190 \\
&= -36 + 72 + 190 \\
&= 226 \text{ feet}.
\end{aligned}
$$[/tex]
Thus, the maximum height of the projectile is [tex]$\boxed{226 \text{ feet}}$[/tex].
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