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. **Identify the Variables:**
- Let \( h \) be the height of the table.
- Let \( x \) be the horizontal distance from the edge of the table where the marble lands.
- Let \( u \) be the horizontal velocity of the marble (which we need to find).
- Let \( g \) be the acceleration due to gravity.

2. **Determine the Time of Flight:**
The marble is projected horizontally, so its initial vertical velocity is zero. The time of flight \( t \) can be derived from the equation of motion for vertical displacement:
\[ h = \frac{1}{2} g t^2 \]
Rearranging this equation to solve for \( t \):
\[ t^2 = \frac{2h}{g} \quad \Rightarrow \quad t = \sqrt{\frac{2h}{g}} \]

3. **Relate Horizontal Distance to Time of Flight:**
The horizontal distance \( x \) traveled by the marble can be expressed as:
\[ x = u \cdot t \]
Substituting the expression for \( t \) from step 2:
\[ x = u \cdot \sqrt{\frac{2h}{g}} \]

4. **Solve for the Horizontal Velocity \( u \):**
Rearranging the equation to solve for \( u \):
\[ u = \frac{x}{\sqrt{\frac{2h}{g}}} \]
Simplifying this expression:
\[ u = x \cdot \sqrt{\frac{g}{2h}} \]

5. **Final Result:**
The velocity of projection \( u \) is given by:
\[ u = x \cdot \sqrt{\frac{g}{2h}} \]

### Final Answer:
The velocity of projection \( u \) is \( x \cdot \sqrt{\frac{g}{2h}} \).