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:

Question

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

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.
make this like a 6th grader wrote it

Answer 1

To understand how high a volleyball goes while it’s in the air, we can use a special math formula called the projectile motion equation. This formula helps us figure out the height \( h \) of the volleyball depending on how long it has been flying, which we call time \( t \). The height is given by:

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

In this formula:

• \( v \) is how fast the volleyball was thrown (in feet per second),
• \( h_0 \) is how high it started (in feet),
• The number \(-16\) is there because of gravity, which pulls the volleyball down to the ground.

For our volleyball, we know:

• It starts at a height of \( h_0 = 6 \) feet (like when it's being held up).
• It’s thrown with a speed of \( v = 33 \) feet per second.

So if we plug these numbers into the formula, we get:

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

Next, we want to figure out when the volleyball reaches its highest point. To do this, we can use a special formula to find the peak of the curve (which is called the vertex). The formula to find that time \( t \) is:

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

In our equation:

• \( a = -16 \)
• \( b = 33 \)

Now let’s do the calculation:

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

That means the volleyball reaches its highest point at about \( \frac{33}{32} \) seconds, which is a little over 1 second.

Now, let’s find out how high the volleyball is when it reaches that moment. We can substitute \( t = \frac{33}{32} \) back into our height equation:

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

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

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

Now we can use that in our height equation:

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

Let’s break it down:

1. First, calculate the part with \(-16\):

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

2. Now, let’s do the next part:

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

3. And then we convert \(6\) into a fraction:

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

Now, we put these parts together:

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

Now let’s combine them:

\[ = \frac{-17424 + 34656 + 6144}{1024} \]

When we simplify that:

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

So, the highest point the volleyball reaches is at:

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

This means the volleyball goes up to exactly 6 feet high and does this about \( 1.03 \) seconds after being thrown.

In simpler words, the volleyball reaches its maximum height of 6 feet after a little over 1 second!