If R is the maximum horizontal range of a particle, then the greatest height attained by it is :

Options

A. R
B. 2R
C. `(R)/(2)`
D. `(R)/(4)`

Explanation

To solve the problem, we need to find the relationship between the maximum horizontal range \( R \) of a projectile and the greatest height \( h \) attained by it.

### Step-by-Step Solution:

1. **Understand the Projectile Motion**:
- A projectile is launched with an initial velocity \( u \) at an angle \( \theta \) with the horizontal. The motion can be analyzed in two dimensions: horizontal (x-axis) and vertical (y-axis).

2. **Maximum Horizontal Range**:
- The formula for the maximum horizontal range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- For maximum range, the angle \( \theta \) is \( 45^\circ \). Thus, \( \sin 2\theta = \sin 90^\circ = 1 \).
- Therefore, the maximum range becomes:
\[ R = \frac{u^2}{g} \]

3. **Greatest Height**:
- The formula for the maximum height \( h \) attained by the projectile is given by:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
- For \( \theta = 45^\circ \), we have \( \sin 45^\circ = \frac{1}{\sqrt{2}} \). Thus, \( \sin^2 45^\circ = \frac{1}{2} \).
- Substituting this into the height formula gives:
\[ h = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

4. **Relating Height and Range**:
- Now, we have:
- Maximum Range: \( R = \frac{u^2}{g} \)
- Maximum Height: \( h = \frac{u^2}{4g} \)
- To express \( h \) in terms of \( R \), we can substitute \( u^2 \) from the range equation into the height equation:
\[ u^2 = R \cdot g \]
- Substituting this into the height equation:
\[ h = \frac{R \cdot g}{4g} = \frac{R}{4} \]

5. **Final Answer**:
- Therefore, the greatest height \( h \) attained by the projectile is:
\[ h = \frac{R}{4} \]

### Conclusion:
The correct answer is \( h = \frac{R}{4} \).