To model the height of the volleyball over time, you can use the equation of a projectile motion. The general form of the height \( h \) as a function of time \( t \) can be given by the formula:

Question

\[
h(t) = -16t^2 + vt + h_0
\]

where:
- \( v \) is the initial velocity (in feet per second),
- \( h_0 \) is the initial height (in feet),
- and the coefficient \(-16\) represents the effect of gravity in feet per second squared.

Given:
- Initial height \( h_0 = 6 \) feet
- Initial velocity \( v = 33 \) feet per second

Substituting these values into the equation:

\[
h(t) = -16t^2 + 33t + 6
\]

To find the vertex of the parabola represented by this equation, we can use the formula for the vertex \( t = -\frac{b}{2a} \) where \( a \) and \( b \) are the coefficients from the standard form \( at^2 + bt + c \).

Here:
- \( a = -16 \)
- \( b = 33 \)

Now substituting these values:

\[
t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds}
\]

Now we can find the height at this time:

\[
h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6
\]

Calculating \( \left(\frac{33}{32}\right)^2 \):

\[
\left(\frac{33}{32}\right)^2 = \frac{1089}{1024}
\]

Substituting back into the height equation:

\[
h\left(\frac{33}{32}\right) = -16 \cdot \frac{1089}{1024} + 33 \cdot \frac{33}{32} + 6
\]
\[
h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{1089}{32} + 6
\]
\[
= -\frac{17424}{1024} + \frac{34656}{1024} + \frac{6144}{1024}
\]
\[
= \frac{17424 - 17424 + 6144}{1024}
\]
\[
= \frac{6144}{1024} = 6
\]

As a result, the vertex of the function is:

\[
\left(\frac{33}{32}, 6\right)
\]

In conclusion, the vertex of the function representing the height of the volleyball is at:

\[
\left(\frac{33}{32}, 6\right) \approx (1.03, 6) \text{ (in feet)}
\]

This means the maximum height reached is 6 feet at approximately \( t = 1.03 \) seconds.
make this like a 9th grader wrote this

Answer 1

To understand how high the volleyball goes over time, we can use a formula from physics called the projectile motion equation. This equation helps us find the height \( h \) of the volleyball as it flies through the air, depending on time \( t \). The height can be expressed with this formula:

\[ h(t) = -16t^2 + vt + h_0 \]

In this formula:

• \( v \) is the initial speed (in feet per second),
• \( h_0 \) is the starting height (in feet),
• The number \(-16\) accounts for gravity, which pulls the volleyball down.

For our problem, we know that:

• The volleyball starts at a height of \( h_0 = 6 \) feet.
• It has an initial speed of \( v = 33 \) feet per second.

If we put these values into our formula, we get:

\[ h(t) = -16t^2 + 33t + 6 \]

Next, we want to find out when the volleyball reaches its highest point. To do this, we can use a special formula to find the vertex of the parabola, which shows us where that maximum height is. The formula for that time \( t \) is:

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

In our equation:

• \( a = -16 \)
• \( b = 33 \)

Plugging in these numbers, we calculate:

\[ t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds} \]

Now, we’ll use this time to find out how high the volleyball is at that moment. We can put \( t = \frac{33}{32} \) back into the height equation:

\[ h\left(\frac{33}{32}\right) = -16\left(\frac{33}{32}\right)^2 + 33\left(\frac{33}{32}\right) + 6 \]

First, let's calculate \( \left(\frac{33}{32}\right)^2 \):

\[ \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \]

Now we plug that back into the height equation:

\[ h\left(\frac{33}{32}\right) = -16 \cdot \frac{1089}{1024} + 33 \cdot \frac{33}{32} + 6 \]

First, calculate each part:

\[ -16 \cdot \frac{1089}{1024} = -\frac{17424}{1024} \]

For the next term:

\[ 33 \cdot \frac{33}{32} = \frac{1089}{32} = \frac{34656}{1024} \]

And \(6\) as:

\[ 6 = \frac{6144}{1024} \]

Now let’s combine these results:

\[ h\left(\frac{33}{32}\right) = -\frac{17424}{1024} + \frac{34656}{1024} + \frac{6144}{1024} \] \[ = \frac{-17424 + 34656 + 6144}{1024} \] \[ = \frac{6144}{1024} = 6 \]

So, the highest point the volleyball reaches is at:

\[ \left(\frac{33}{32}, 6\right) \]

This is about:

\[ (1.03, 6) \text{ feet} \]

In simpler terms, this means the volleyball, at its maximum height, reaches exactly 6 feet after approximately \( t = 1.03 \) seconds.