Water from a tap emerges vertically downwards with an initial velocity `V_(0)`. Assume pressure is constant throughout the stream of water and the flow is steady. Find the distance form the tap at which cross-sectional area of stream is half of the cross-sectional area of stream at the tap.

Options

A. `V_(0)^(2)//2g`
B. `3V_(0)^(2)//2g`
C. `2V_(0)^(2)//g`
D. `5V_(0)^(2)//2g`

Explanation

To solve the problem step by step, we will use the principles of fluid mechanics, specifically the equations of motion and the continuity equation.

### Step 1: Understand the Problem
Water is flowing from a tap vertically downward with an initial velocity \( V_0 \). We need to find the distance \( h \) from the tap where the cross-sectional area of the stream is half of the area at the tap.

### Step 2: Use the Continuity Equation
The continuity equation states that the product of the cross-sectional area and the velocity of the fluid must remain constant along the flow. Mathematically, this is expressed as:
\[ A_1 V_1 = A_2 V_2 \]
Where:
- \( A_1 \) = cross-sectional area at the tap
- \( V_1 \) = initial velocity \( V_0 \)
- \( A_2 \) = cross-sectional area at distance \( h \)
- \( V_2 \) = velocity at distance \( h \)

Given that \( A_2 = \frac{1}{2} A_1 \), we can substitute this into the continuity equation:
\[ A_1 V_0 = \left(\frac{1}{2} A_1\right) V_2 \]
This simplifies to:
\[ V_2 = 2 V_0 \]

### Step 3: Use the Equation of Motion
Next, we apply the equation of motion to find the relationship between the velocity and the height. The relevant equation is:
\[ V_2^2 = V_0^2 + 2gh \]
Substituting \( V_2 = 2 V_0 \) into this equation gives:
\[ (2 V_0)^2 = V_0^2 + 2gh \]
This simplifies to:
\[ 4 V_0^2 = V_0^2 + 2gh \]

### Step 4: Rearranging the Equation
Now, we can rearrange the equation to solve for \( h \):
\[ 4 V_0^2 - V_0^2 = 2gh \]
\[ 3 V_0^2 = 2gh \]
\[ h = \frac{3 V_0^2}{2g} \]

### Final Answer
The distance from the tap at which the cross-sectional area of the stream is half of the cross-sectional area at the tap is:
\[ h = \frac{3 V_0^2}{2g} \]

---