Question

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!

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Answer 1

To determine the maximum height of a volleyball in the air, we use the projectile motion equation:

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

where:
- \( v = 33 \) feet per second (initial speed),
- \( h_0 = 6 \) feet (initial height).

Plugging in these values gives:

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

To find the time \( t \) when the volleyball reaches its highest point, we use the vertex formula:

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

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

\[
t = -\frac{33}{2 \times -16} = \frac{33}{32} \text{ seconds} \approx 1.03 \text{ seconds}
\]

Next, we calculate 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 \( \left(\frac{33}{32}\right)^2 = \frac{1089}{1024} \):

Now substituting into the height equation:

1. \(-16 \cdot \frac{1089}{1024} = -\frac{17424}{1024}\)
2. \(33 \cdot \frac{33}{32} = \frac{1089}{32} = \frac{34656}{1024}\)
3. \(6 = \frac{6144}{1024}\)

Putting it all together:

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

Thus, the volleyball reaches a maximum height of **6 feet** at approximately **1.03 seconds** after being thrown.