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To solve this problem, we need to determine the value of \( h \), the height of the cylindrical part of the container, based on the information provided. Here’s the step-by-step approach:

### Step 1: Understanding the Container Dimensions
1. The container consists of:
   - A cone with radius \( r = 3 \) m and height \( 4 \) m.
   - A cylinder with the same radius \( r = 3 \) m and unknown height \( h \) m.

### Step 2: Volumes of the Cone and Cylinder
The volume of a cone and cylinder can be calculated using the formulas:
- **Volume of a cone:** \( V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} \)
- **Volume of a cylinder:** \( V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} \)

Substitute \( r = 3 \) m and \( h_{\text{cone}} = 4 \) m:
1. **Volume of the cone:**
   \[
   V_{\text{cone}} = \frac{1}{3} \pi (3)^2 \cdot 4 = \frac{1}{3} \pi \cdot 9 \cdot 4 = 12 \pi \text{ cubic meters}
   \]
2. **Volume of the cylinder (with height \( h \)):**
   \[
   V_{\text{cylinder}} = \pi (3)^2 \cdot h = 9\pi h \text{ cubic meters}
   \]

### Step 3: Total Volume of the Container
The total volume of the container, when full, is the sum of the volumes of the cone and the cylinder:
\[
V_{\text{total}} = V_{\text{cone}} + V_{\text{cylinder}} = 12 \pi + 9 \pi h
\]

### Step 4: Determine the Rate of Filling
From the problem, we know:
- After 5 hours, the depth of the grain is 6 meters above the vertex of the cone.
- This depth includes the entire 4 meters of the cone and an additional 2 meters into the cylinder.

Thus, in 5 hours, the grain has filled:
1. The entire cone volume: \( 12 \pi \)
2. Plus the cylinder section up to a height of 2 meters: \( 9 \pi \cdot 2 = 18 \pi \)

So, the volume filled in 5 hours is:
\[
12 \pi + 18 \pi = 30 \pi
\]

Since this volume was filled in 5 hours, the rate of filling is:
\[
\frac{30 \pi}{5} = 6 \pi \text{ cubic meters per hour}
\]

### Step 5: Total Time to Fill the Container
The container is completely full after 9 hours, so the total volume filled in that time is:
\[
9 \text{ hours} \times 6 \pi = 54 \pi \text{ cubic meters}
\]

### Step 6: Setting Up the Equation
Since the total volume when the container is full is \( V_{\text{total}} = 12 \pi + 9 \pi h \), we equate this to \( 54 \pi \):
\[
12 \pi + 9 \pi h = 54 \pi
\]

### Step 7: Solving for \( h \)
1. Divide both sides by \( \pi \):
   \[
   12 + 9h = 54
   \]
2. Subtract 12 from both sides:
   \[
   9h = 42
   \]
3. Divide by 9:
   \[
   h = \frac{42}{9} = \frac{14}{3}
   \]

### Final Answer
The value of \( h \) is:
\[
h = \frac{14}{3} \text{ meters}
\]

Would you like further details on any step or have any other questions?

---

### Related Questions
1. How do you calculate the volume of a cone or cylinder in different units?
2. How does the filling rate affect the volume calculations for variable-depth containers?
3. What would be the impact on \( h \) if the cylinder had a different radius?
4. How would this problem change if the cone's height was altered but the cylinder’s height remained constant?
5. How can you verify the filling rate through alternative methods?

### Tip
When working with composite shapes, breaking down the problem by calculating individual volumes and understanding rates can simplify complex volumetric problems.

The container is a cylinder on top of a cone. The cylinder has a radius of 3 m and a height of h m. The cone has a base radius of 3 m and a vertical height of 4 m. The container is empty and is filled with grain at a constant rate. After 5 hours, the depth of the grain is 6 meters above the vertex of the cone, and after 9 hours, the container is full. Work out the value of h. Give your answer as a fraction in its simplest form and show all workings.

Learn how to determine the height of a cylindrical container on top of a cone, filled at a constant rate. This problem involves calculating the volume of a cone and cylinder and understanding rates of change. Perfect for students in Grades 10-12.

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