Question

A flare was launched straight up from the ground with an initial velocity of 176 ft/s and returned to the ground after 11 s.

The height of the flare t seconds after launch is modeled by the function f(t)=−16t2+176t.

What is the maximum height of the flare, in feet?

Enter your answer in the box.

Answer 1

To find the maximum height of the flare, we need to determine the vertex of the quadratic function given by \( f(t) = -16t^2 + 176t \). The maximum height occurs at the vertex since the parabola opens downwards (as indicated by the negative coefficient of \( t^2 \)).

The formula for the time at which the vertex occurs, for a quadratic function of the form \( f(t) = at^2 + bt + c \), is given by:

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

Here, \( a = -16 \) and \( b = 176 \). Plugging in these values:

\[
t = -\frac{176}{2 \times -16} = \frac{176}{32} = 5.5 \text{ seconds}
\]

Now, we can substitute \( t = 5.5 \) back into the height function \( f(t) \) to find the maximum height:

\[
f(5.5) = -16(5.5^2) + 176(5.5)
\]

Calculating \( 5.5^2 \):

\[
5.5^2 = 30.25
\]

Now substitute \( 30.25 \) back into the equation:

\[
f(5.5) = -16(30.25) + 176(5.5)
\]

Calculating \( -16 \times 30.25 \):

\[
-16 \times 30.25 = -484
\]

Now calculate \( 176 \times 5.5 \):

\[
176 \times 5.5 = 968
\]

Now combine the two results:

\[
f(5.5) = -484 + 968 = 484
\]

Thus, the maximum height of the flare is:

\[
\boxed{484} \text{ feet}
\]