A cylinderical vessel is filled with water up to height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range h is equal to

Options

A. `H//4`
B. `H//2`
C. `3H//4`
D. `H`

Explanation

To solve the problem of finding the depth \( h \) from the free surface of water for maximum range when water flows out of a hole in a cylindrical vessel, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters**:
- Let \( H \) be the height of the water in the cylindrical vessel.
- Let \( h \) be the depth of the hole from the free surface of the water.

2. **Determine the velocity of water exiting the hole**:
- The velocity \( U \) of water exiting the hole can be calculated using Torricelli's theorem:
\[ U = \sqrt{2gh} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Calculate the time of flight**:
- The water will fall a vertical distance of \( H - h \) before it reaches the ground. Using the equation of motion:
\[ S_y = U_y t + \frac{1}{2} g t^2 \]
- Since the initial vertical velocity \( U_y = 0 \):
\[ H - h = \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ t^2 = \frac{2(H - h)}{g} \]
- Thus, the time \( t \) is:
\[ t = \sqrt{\frac{2(H - h)}{g}} \]

4. **Calculate the horizontal range**:
- The horizontal range \( R \) can be calculated as:
\[ R = U \cdot t \]
- Substituting for \( U \) and \( t \):
\[ R = \sqrt{2gh} \cdot \sqrt{\frac{2(H - h)}{g}} = 2\sqrt{h(H - h)} \]

5. **Maximize the range**:
- To find the maximum range, we need to differentiate \( R \) with respect to \( h \) and set the derivative to zero:
\[ R = 2\sqrt{h(H - h)} \]
- Let \( R^2 = 4h(H - h) \). Differentiate \( R^2 \):
\[ \frac{d(R^2)}{dh} = 4(H - 2h) = 0 \]
- Setting the derivative to zero gives:
\[ H - 2h = 0 \implies h = \frac{H}{2} \]

6. **Conclusion**:
- For maximum range, the depth \( h \) from the free surface of water should be:
\[ h = \frac{H}{2} \]

### Final Answer:
The value of \( h \) for maximum range is \( \frac{H}{2} \).