A glass marble projected horizontally from the top of a table falls at a distance x from the edge of the table. If h is the height of the table, then the velocity of projection is

Options

A. `h sqrt (g/(2x))`
B. `x sqrt(g/(2h))`
C. gxh
D. gx +h

Explanation

To solve the problem of finding the velocity of projection of a glass marble projected horizontally from the top of a table, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
- A marble is projected horizontally from the edge of a table of height \( h \).
- It falls a horizontal distance \( x \) from the edge of the table before hitting the ground.

2. **Identify the Variables**:
- Let \( u \) be the initial horizontal velocity of the marble (which we need to find).
- The height of the table is \( h \).
- The horizontal distance traveled by the marble is \( x \).
- The acceleration due to gravity is \( g \).

3. **Determine the Time of Flight**:
- The time \( t \) it takes for the marble to fall to the ground can be determined using the equation of motion for vertical displacement:
\[ h = \frac{1}{2} g t^2 \]
- Rearranging this gives:
\[ t^2 = \frac{2h}{g} \]
- Taking the square root gives:
\[ t = \sqrt{\frac{2h}{g}} \]

4. **Relate Horizontal Distance and Time**:
- Since there is no acceleration in the horizontal direction, the horizontal motion is uniform. The horizontal distance \( x \) is given by:
\[ x = u \cdot t \]
- Rearranging this for \( u \) gives:
\[ u = \frac{x}{t} \]

5. **Substitute the Time of Flight**:
- Substitute the expression for \( t \) from step 3 into the equation for \( u \):
\[ u = \frac{x}{\sqrt{\frac{2h}{g}}} \]
- This simplifies to:
\[ u = x \cdot \sqrt{\frac{g}{2h}} \]

6. **Final Expression**:
- Therefore, the velocity of projection \( u \) is:
\[ u = x \cdot \sqrt{\frac{g}{2h}} \]

### Final Answer:
The velocity of projection \( u \) is given by:
\[ u = x \sqrt{\frac{g}{2h}} \]