Question

Austin launches a toy rocket from a platform. The height of the rocket in feet is given by h, of, t, equals, minus, 16, t, squared, plus, 128, t, plus, 68h(t)=−16t
2
+128t+68 where tt represents the time in seconds after launch. What is the rocket’s greatest height?

Answer 1

To find the rocket's greatest height, we need to determine the maximum value of the quadratic function given by:

\[ h(t) = -16t^2 + 128t + 68. \]

This is a downward-opening parabola (since the coefficient of \(t^2\) is negative). The maximum height occurs at the vertex of the parabola.

The time \( t \) at which the maximum height occurs can be found using the formula:

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

where \( a \) is the coefficient of \( t^2 \) and \( b \) is the coefficient of \( t \).

In our function:
- \( a = -16 \)
- \( b = 128 \)

Now we can plug these values into the vertex formula:

\[
t = -\frac{128}{2 \times -16} = -\frac{128}{-32} = 4 \, \text{seconds}.
\]

Next, we substitute \( t = 4 \) back into the height function \( h(t) \) to find the maximum height:

\[
h(4) = -16(4^2) + 128(4) + 68.
\]

Calculating \( h(4) \):

\[
h(4) = -16(16) + 128(4) + 68,
\]

\[
h(4) = -256 + 512 + 68,
\]

\[
h(4) = -256 + 580 = 324.
\]

Therefore, the rocket's greatest height is:

\[
\text{Greatest Height} = 324 \, \text{feet}.
\]