Question

A football player kicks the ball with an initial upward velocity of 60 feet per second. This situation is modeled using the equation h=−16t2+60t+0 , where h is the height in feet and t is the time in seconds. Using this equation, what is the maximum range the ball will reach? Round to the nearest hundredth.(1 point) ()feet

Answer 1

To find the maximum height the ball reaches, we can use the vertex form of a quadratic function. The general form of the equation you provided is:

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

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

- \(a = -16\)
- \(b = 60\)
- \(c = 0\)

The maximum height occurs at the vertex of the parabola. For a quadratic function of the form \(h(t) = at^2 + bt + c\), the time \(t\) at which the maximum height occurs can be calculated using the formula:

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

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

\[
t = -\frac{60}{2 \times -16} = -\frac{60}{-32} = \frac{60}{32} = 1.875 \text{ seconds}
\]

Next, we substitute \(t = 1.875\) back into the height equation to find the maximum height \(h(t)\):

\[
h(1.875) = -16(1.875)^2 + 60(1.875) + 0
\]

Calculating \( (1.875)^2 \):

\[
(1.875)^2 = 3.515625
\]

Now substituting this back into the equation:

\[
h(1.875) = -16(3.515625) + 60(1.875)
\]

Calculating each term:

1. \(-16 \times 3.515625 = -56.25\)
2. \(60 \times 1.875 = 112.5\)

Now, combine them:

\[
h(1.875) = -56.25 + 112.5 = 56.25 \text{ feet}
\]

Thus, the maximum height the ball will reach is:

\[
\boxed{56.25} \text{ feet}
\]