Question

Indigo 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, 80, t, plus, 96h(t)=−16t
2
+80t+96 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 identify the maximum point of the quadratic function given by the height equation:

\[
h(t) = -16t^2 + 80t + 96
\]

This is a quadratic function in the standard form \(h(t) = at^2 + bt + c\), where:

- \(a = -16\)
- \(b = 80\)
- \(c = 96\)

Since \(a\) is negative, the parabola opens downwards, and the maximum height occurs at the vertex of the parabola.

The time at which the maximum height occurs is given by the formula:

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

Substituting the values of \(a\) and \(b\):

\[
t = -\frac{80}{2(-16)} = -\frac{80}{-32} = 2.5 \text{ seconds}
\]

Now, we can find the maximum height \(h(t)\) by substituting \(t = 2.5\) back into the height equation:

\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 96
\]
\[
= -16(6.25) + 80(2.5) + 96
\]
\[
= -100 + 200 + 96
\]
\[
= 196 \text{ feet}
\]

Thus, the rocket's greatest height is **196 feet**.