A wide cylindrical vessel `50cm` in height is filled with water and rests on a table. Assuming the viscosity to be negligible, find at what height from the bottom of the vessel a small hole should be perforated for the water jet coming out of it to hit the surface of the table at the maximum distance `l_(max)` from the vessel. Find `l_(max)`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a wide cylindrical vessel filled with water, and we need to determine the height \( h \) from the bottom of the vessel at which a small hole should be made so that the water jet coming out of it hits the table at the maximum distance \( L_{\text{max}} \).

### Step 2: Apply Bernoulli's Theorem
According to Bernoulli's theorem, the energy per unit volume is conserved. For the water jet, we can equate the potential energy at height \( H - h \) to the kinetic energy of the water jet:

\[ \frac{1}{2} \rho v^2 = \rho g (H - h) \]

Where:
- \( \rho \) is the density of water,
- \( v \) is the velocity of water coming out of the hole,
- \( g \) is the acceleration due to gravity,
- \( H \) is the total height of the water (50 cm),
- \( h \) is the height from the bottom of the vessel where the hole is made.

### Step 3: Solve for Velocity \( v \)
From the equation above, we can solve for \( v \):

\[ v = \sqrt{2g(H - h)} \]

### Step 4: Determine the Time of Flight
The water will fall a vertical distance \( h \) before hitting the table. The time \( t \) taken to fall this distance can be calculated using the equation of motion:

\[ h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}} \]

### Step 5: Calculate the Horizontal Range \( L \)
The horizontal range \( L \) where the water hits the table can be calculated using the formula:

\[ L = v \cdot t \]

Substituting the values of \( v \) and \( t \):

\[ L = \sqrt{2g(H - h)} \cdot \sqrt{\frac{2h}{g}} = 2\sqrt{h(H - h)} \]

### Step 6: Maximize the Range \( L \)
To find the height \( h \) that maximizes \( L \), we differentiate \( L \) with respect to \( h \) and set the derivative to zero:

\[ \frac{dL}{dh} = 2\left(\frac{1}{2\sqrt{h(H - h)}}(H - h) - \frac{1}{2\sqrt{h(H - h)}}h\right) = 0 \]

This simplifies to:

\[ H - 2h = 0 \implies h = \frac{H}{2} \]

### Step 7: Calculate \( h \) and \( L_{\text{max}} \)
Given \( H = 50 \, \text{cm} \):

\[ h = \frac{50}{2} = 25 \, \text{cm} \]

Now, substituting \( h = 25 \, \text{cm} \) back into the equation for \( L \):

\[ L_{\text{max}} = 2\sqrt{25(50 - 25)} = 2\sqrt{25 \cdot 25} = 2 \cdot 25 = 50 \, \text{cm} \]

### Final Answers
- The height \( h \) from the bottom of the vessel should be **25 cm**.
- The maximum distance \( L_{\text{max}} \) is **50 cm**.

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