An object is projected horizontally from a top of the tower of height h. The line joining the point of projection and point of striking on the ground makes an angle `45^@` with ground, then with what velocity the object strikes the ground 

Options

A. `sqrt((11gh)/(2))`
B. `sqrt((9gh)/(2))`
C. `sqrt((7gh)/(2))`
D. `sqrt((5gh)/(2))`

Answer

To solve the problem, we need to find the velocity with which the object strikes the ground after being projected horizontally from the top of a tower of height \( h \). The line joining the point of projection and the point of striking on the ground makes an angle of \( 45^\circ \) with the ground.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- An object is projected horizontally from a height \( h \).
- The angle between the line of projection and the ground at the point of impact is \( 45^\circ \).

2. **Determine the Time of Flight**:
- The time \( t \) taken to fall from height \( h \) can be calculated using the formula for free fall:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ t = \sqrt{\frac{2h}{g}} \]

3. **Horizontal Motion**:
- Let \( u \) be the horizontal velocity with which the object is projected.
- The horizontal distance \( x \) covered during time \( t \) is:
\[ x = u \cdot t = u \cdot \sqrt{\frac{2h}{g}} \]

4. **Using the Angle of Impact**:
- Since the angle of impact is \( 45^\circ \), we know that the horizontal and vertical components of velocity at the point of impact are equal.
- The vertical component of velocity \( v_y \) just before striking the ground can be calculated using:
\[ v_y = g t = g \cdot \sqrt{\frac{2h}{g}} = \sqrt{2gh} \]
- The horizontal component of velocity \( v_x \) is equal to the initial horizontal velocity \( u \).

5. **Setting Up the Equation**:
- Since the angle is \( 45^\circ \), we have:
\[ v_y = v_x \]
- Therefore:
\[ \sqrt{2gh} = u \]

6. **Finding the Resultant Velocity**:
- The resultant velocity \( v \) when the object strikes the ground can be calculated using Pythagoras' theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + (2gh)} = \sqrt{u^2 + 2gh} \]
- Substituting \( u = \sqrt{2gh} \):
\[ v = \sqrt{(\sqrt{2gh})^2 + 2gh} = \sqrt{2gh + 2gh} = \sqrt{4gh} = 2\sqrt{gh} \]

7. **Final Velocity**:
- To find the final expression, we need to consider the total energy or kinematic equations. Using the kinematic equation:
\[ v^2 = u^2 + 2gh \]
- Since we have \( u^2 = \frac{gh}{2} \):
\[ v^2 = \frac{gh}{2} + 2gh = \frac{gh}{2} + \frac{4gh}{2} = \frac{5gh}{2} \]
- Thus:
\[ v = \sqrt{\frac{5gh}{2}} \]

### Conclusion:
The velocity with which the object strikes the ground is:
\[ v = \sqrt{\frac{5gh}{2}} \]