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Determining the Vertex

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation [tex]h(t)=-16t^2+48t+190[/tex].

What is the maximum height of the projectile?

A. 82 feet
B. 190 feet
C. 226 feet
D. 250 feet

Answer :

To find the maximum height of the projectile, we need to determine the vertex of the quadratic equation [tex]h(t) = -16t^2 + 48t + 190[/tex], which represents the height of the projectile over time. The vertex of a parabola in the form [tex]ax^2 + bx + c[/tex] is given by the formula:

[tex]t = \frac{-b}{2a}[/tex]

Here, [tex]a = -16[/tex] and [tex]b = 48[/tex]. Plug these values into the formula:

[tex]t = \frac{-48}{2(-16)} = \frac{-48}{-32} = 1.5[/tex]

The vertex occurs at [tex]t = 1.5[/tex] seconds. To find the maximum height (the [tex]h[/tex] value at the vertex), we substitute [tex]t = 1.5[/tex] back into the original equation:

[tex]h(1.5) = -16(1.5)^2 + 48(1.5) + 190[/tex]

Now calculate each term:

  1. [tex]-16(1.5)^2 = -16(2.25) = -36[/tex]
  2. [tex]48(1.5) = 72[/tex]
  3. Constant term is [tex]+190[/tex]

Adding these values gives:

[tex]h(1.5) = -36 + 72 + 190 = 226[/tex]

Thus, the maximum height of the projectile is 226 feet.

Therefore, the correct answer is C. 226 feet.

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