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.