Question

A football is kicked with an initial upward velocity of 60 feet per second. It can be modeled by the equation h= -16t^2+60t+0. Where h is the height in feet and t is the time in seconds. What is the maximum range the ball will reach

Answer 1

To find the maximum height that the football reaches, we can analyze the given quadratic equation for height, which is:

\[
h(t) = -16t^2 + 60t + 0
\]

This equation is in the standard form \(h(t) = at^2 + bt + c\), where:
- \(a = -16\)
- \(b = 60\)
- \(c = 0\)

The maximum height of a quadratic function occurs at the vertex. The time \(t\) at which this maximum occurs can be found using the formula:

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

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

\[
t = -\frac{60}{2 \times -16} = -\frac{60}{-32} = \frac{60}{32} = \frac{15}{8} \text{ seconds}
\]

Now we can find the maximum height by substituting this \(t\) value back into the height equation:

\[
h\left(\frac{15}{8}\right) = -16\left(\frac{15}{8}\right)^2 + 60\left(\frac{15}{8}\right) + 0
\]

Calculating \(\left(\frac{15}{8}\right)^2\):

\[
\left(\frac{15}{8}\right)^2 = \frac{225}{64}
\]

Now substituting back into the height equation:

\[
h\left(\frac{15}{8}\right) = -16\left(\frac{225}{64}\right) + 60\left(\frac{15}{8}\right)
\]

Calculating each term:

1. \(-16\left(\frac{225}{64}\right) = -\frac{3600}{64} = -56.25\)
2. \(60\left(\frac{15}{8}\right) = \frac{900}{8} = 112.5\)

Now add these two results together:

\[
h\left(\frac{15}{8}\right) = -56.25 + 112.5 = 56.25 \text{ feet}
\]

Thus, the **maximum height** that the football will reach is **56.25 feet**.