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Answer :
To find the maximum height of a projectile, we need to understand that the given equation [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex] represents the height of the projectile over time [tex]\( t \)[/tex]. This is a quadratic equation of the form [tex]\( at^2 + bt + c \)[/tex].
The path of the projectile forms a parabola, and since the coefficient of [tex]\( t^2 \)[/tex] is negative ([tex]\(-16\)[/tex]), the parabola opens downward. This means it has a maximum point, which is at its vertex.
The formula to find the time at which the vertex (maximum point) of a parabola occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For our equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Plug these into the formula to find [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \][/tex]
This means that the maximum height occurs at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we want to calculate the height at this time by plugging [tex]\( t = 1.5 \)[/tex] back into the original equation:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
[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.
The path of the projectile forms a parabola, and since the coefficient of [tex]\( t^2 \)[/tex] is negative ([tex]\(-16\)[/tex]), the parabola opens downward. This means it has a maximum point, which is at its vertex.
The formula to find the time at which the vertex (maximum point) of a parabola occurs is given by:
[tex]\[ t = -\frac{b}{2a} \][/tex]
For our equation:
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex]
Plug these into the formula to find [tex]\( t \)[/tex]:
[tex]\[ t = -\frac{48}{2 \times -16} = \frac{48}{32} = 1.5 \][/tex]
This means that the maximum height occurs at [tex]\( t = 1.5 \)[/tex] seconds.
Next, we want to calculate the height at this time by plugging [tex]\( t = 1.5 \)[/tex] back into the original equation:
[tex]\[ h(1.5) = -16(1.5)^2 + 48(1.5) + 190 \][/tex]
[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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