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 of finding the greatest height attained by a projectile when the maximum horizontal range \( R \) is given, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to find the relationship between the maximum horizontal range \( R \) and the greatest height \( h \) attained by a projectile.

2. **Use the Range Formula**: The formula for the range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

3. **Determine the Angle for Maximum Range**: The range is maximized when \( \theta = 45^\circ \). Therefore, we can substitute \( \theta = 45^\circ \) into the range formula:
\[ R = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} \]
since \( \sin(90^\circ) = 1 \).

4. **Relate Initial Velocity to Maximum Height**: The formula for the maximum height \( h \) attained by the projectile is given by:
\[ h = \frac{u^2 \sin^2(\theta)}{2g} \]
Again, substituting \( \theta = 45^\circ \):
\[ h = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

5. **Substitute \( u^2 \) from the Range Formula**: From the range formula, we have \( u^2 = Rg \). Substituting this into the height formula:
\[ h = \frac{Rg}{4g} = \frac{R}{4} \]

6. **Final Result**: Therefore, the greatest height \( h \) attained by the projectile when the maximum horizontal range \( R \) is given is:
\[ h = \frac{R}{4} \]

### Summary:
The greatest height attained by a projectile when the maximum horizontal range \( R \) is given is \( \frac{R}{4} \).