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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 a projectile launched from a building, given the equation [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex], we need to follow the process of finding the vertex of this quadratic equation. The vertex of a parabolic equation [tex]\( ax^2 + bx + c \)[/tex] represents either its maximum or minimum point. In this case, since the coefficient of [tex]\( t^2 \)[/tex] is negative, the parabola opens downwards, and the vertex gives the maximum height.

Here are the steps to solve the problem:

1. Identify the coefficients from the equation:
The given equation is [tex]\( h(t) = -16t^2 + 48t + 190 \)[/tex].
- [tex]\( a = -16 \)[/tex]
- [tex]\( b = 48 \)[/tex]
- [tex]\( c = 190 \)[/tex] (this is not required to find the vertex, but it is part of our quadratic equation)

2. Find the time at which the maximum height occurs:
The time [tex]\( t \)[/tex] at which the maximum height occurs can be found using the formula for the vertex of a parabola:

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

Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:

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

3. Calculate the maximum height:
Plug the value of [tex]\( t = 1.5 \)[/tex] back into the original equation to find [tex]\( h(t) \)[/tex]:

[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]

Then, substitute back into the equation:

[tex]\[
h(1.5) = -16 \times 2.25 + 48 \times 1.5 + 190
\][/tex]

Calculate each term:

[tex]\[
-16 \times 2.25 = -36
\][/tex]

[tex]\[
48 \times 1.5 = 72
\][/tex]

Now, combine all terms:

[tex]\[
h(1.5) = -36 + 72 + 190 = 226 \, \text{feet}
\][/tex]

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

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