a body is thrown horizontally with a velocity `sqrt(2gh)` from the top of a tower of height h. It strikes the level gound through the foot of the tower at a distance x from the tower. The value of x is :-

Options

A. h
B. `h//2`
C. `2h`
D. `2h//3`

Explanation

To solve the problem, we need to determine the horizontal distance \( x \) that a body thrown horizontally from the top of a tower with height \( h \) travels before it hits the ground. The body is thrown with an initial horizontal velocity of \( \sqrt{2gh} \).

### Step-by-Step Solution:

1. **Identify the parameters:**
- Height of the tower: \( h \)
- Initial horizontal velocity: \( u = \sqrt{2gh} \)
- Acceleration due to gravity: \( g \)

2. **Calculate the time of flight:**
The body falls a vertical distance \( h \) under the influence of gravity. We can use the second equation of motion for vertical motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = h \), \( u = 0 \) (since the initial vertical velocity is zero), and \( a = g \). Therefore, we have:
\[ h = 0 + \frac{1}{2} g t^2 \]
Simplifying this gives:
\[ h = \frac{1}{2} g t^2 \implies t^2 = \frac{2h}{g} \implies t = \sqrt{\frac{2h}{g}} \]

3. **Calculate the horizontal distance \( x \):**
The horizontal distance traveled by the body is given by:
\[ x = u \cdot t \]
Substituting the values of \( u \) and \( t \):
\[ x = \sqrt{2gh} \cdot \sqrt{\frac{2h}{g}} = \sqrt{2gh} \cdot \sqrt{2h} \cdot \frac{1}{\sqrt{g}} = \sqrt{\frac{4h^2}{g}} = \frac{2h}{\sqrt{g}} \]

4. **Final result:**
Thus, the horizontal distance \( x \) from the foot of the tower to the point where the body strikes the ground is:
\[ x = \frac{2h}{\sqrt{g}} \]

### Conclusion:
The value of \( x \) is \( \frac{2h}{\sqrt{g}} \).