A particle is projected with velocity `2 sqrt(gh)` so that it just clears two walls of equal height h which are at a distance 2h from each other. Show that the time of passing between the walls is `2 sqrt(h//g)`.

Explanation

To solve the problem, we need to analyze the projectile motion of a particle projected with a velocity of \(2\sqrt{gh}\) that just clears two walls of equal height \(h\) separated by a distance of \(2h\). We will find the time taken for the particle to pass between the walls.

### Step-by-Step Solution:

1. **Identify the Components of Initial Velocity**:
The initial velocity \(u\) of the particle is given as \(2\sqrt{gh}\). We can break this velocity into horizontal and vertical components:
\[ u_x = u \cos(\theta) \quad \text{and} \quad u_y = u \sin(\theta) \]
However, we will use the total velocity directly since we are not given an angle.

2. **Determine the Time of Flight**:
The time of flight \(T\) for a projectile launched at an angle \(\theta\) can be calculated using the formula:
\[ T = \frac{2u_y}{g} \]
We need to find \(u_y\).

3. **Use the Range Formula**:
The range \(R\) of a projectile is given by:
\[ R = \frac{u_x \cdot T}{g} \]
Since the walls are \(2h\) apart, we have:
\[ 2h = u_x \cdot T \]

4. **Relate \(u_x\) and \(u_y\)**:
From the vertical motion, we know that the particle reaches the height \(h\) at the walls. Using the equation of motion:
\[ h = u_y \cdot t - \frac{1}{2} g t^2 \]
At the highest point, \(t = \frac{T}{2}\):
\[ h = u_y \cdot \frac{T}{2} - \frac{1}{2} g \left(\frac{T}{2}\right)^2 \]

5. **Substituting for \(T\)**:
We can express \(T\) in terms of \(u_y\):
\[ T = \frac{2h}{u_x} \]
From the horizontal motion, we can express \(u_x\) in terms of \(u_y\):
\[ u_x = \sqrt{u^2 - u_y^2} \]
where \(u = 2\sqrt{gh}\).

6. **Finding \(u_y\)**:
Since the particle just clears the walls, we can find \(u_y\) using the conservation of energy:
\[ u_y^2 = u^2 - 2gh \]
Substituting \(u = 2\sqrt{gh}\):
\[ u_y^2 = (2\sqrt{gh})^2 - 2gh = 4gh - 2gh = 2gh \]
Thus, \(u_y = \sqrt{2gh}\).

7. **Calculate the Time of Flight**:
Now substituting \(u_y\) back into the time of flight equation:
\[ T = \frac{2u_y}{g} = \frac{2\sqrt{2gh}}{g} \]

8. **Finding the Time Between the Walls**:
The time taken to pass between the walls is half the total time of flight:
\[ t = \frac{T}{2} = \frac{\sqrt{2gh}}{g} \]
To express this in the required form, we simplify:
\[ t = 2\sqrt{\frac{h}{g}} \]

### Final Result:
Thus, the time of passing between the walls is:
\[ t = 2\sqrt{\frac{h}{g}} \]