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Answer :
To find the maximum height of the projectile, we need to understand the path described by the quadratic equation [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex]. This equation represents a parabola, and the maximum height is found at the vertex of this parabola.
Here is a step-by-step solution:
1. Determine the Nature of the Parabola:
The parabola opens downwards because the coefficient of [tex]\( t^2 \)[/tex] is negative (-16).
2. Find the Time at Vertex [tex]\( t \)[/tex]:
The formula to find the time [tex]\( t \)[/tex] at which the vertex occurs (and hence the maximum height) for a quadratic equation [tex]\( at^2 + bt + c \)[/tex] is given by:
[tex]\[
t = \frac{-b}{2a}
\][/tex]
In this equation, the values are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
Plugging these values into the formula gives:
[tex]\[
t = \frac{-48}{2 \times -16} = 1.5
\][/tex]
3. Calculate the Maximum Height:
Substitute [tex]\( t = 1.5 \)[/tex] back into the original height equation to find the maximum height [tex]\( h(t) \)[/tex]:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Solving this:
[tex]\[
h(1.5) = -16(2.25) + 72 + 190
\][/tex]
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
Therefore, the maximum height of the projectile is 226 feet.
Here is a step-by-step solution:
1. Determine the Nature of the Parabola:
The parabola opens downwards because the coefficient of [tex]\( t^2 \)[/tex] is negative (-16).
2. Find the Time at Vertex [tex]\( t \)[/tex]:
The formula to find the time [tex]\( t \)[/tex] at which the vertex occurs (and hence the maximum height) for a quadratic equation [tex]\( at^2 + bt + c \)[/tex] is given by:
[tex]\[
t = \frac{-b}{2a}
\][/tex]
In this equation, the values are:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
Plugging these values into the formula gives:
[tex]\[
t = \frac{-48}{2 \times -16} = 1.5
\][/tex]
3. Calculate the Maximum Height:
Substitute [tex]\( t = 1.5 \)[/tex] back into the original height equation to find the maximum height [tex]\( h(t) \)[/tex]:
[tex]\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 190
\][/tex]
Solving this:
[tex]\[
h(1.5) = -16(2.25) + 72 + 190
\][/tex]
[tex]\[
h(1.5) = -36 + 72 + 190
\][/tex]
[tex]\[
h(1.5) = 226
\][/tex]
Therefore, the maximum height of the projectile is 226 feet.
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