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. `(3)/(4)h`

Answer

To solve the problem, we need to determine the horizontal distance \( x \) that the body travels before it hits the ground after being thrown horizontally from the top of a tower of height \( h \) with an initial 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 time \( t \) it takes for the body to fall from the height \( h \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
Rearranging the formula to solve for \( t \):
\[ t^2 = \frac{2h}{g} \]
Taking the square root of both sides gives:
\[ t = \sqrt{\frac{2h}{g}} \]

3. **Calculate the horizontal distance \( x \)**:
The horizontal distance \( x \) traveled by the body can be calculated using the formula:
\[ x = u \cdot t \]
Substituting the values of \( u \) and \( t \):
\[ x = \sqrt{2gh} \cdot \sqrt{\frac{2h}{g}} \]
Simplifying this expression:
\[ x = \sqrt{2gh} \cdot \sqrt{\frac{2h}{g}} = \sqrt{(2h)(2h)} = \sqrt{4h^2} = 2h \]

4. **Final answer**:
Thus, the value of \( x \) is:
\[ x = 2h \]