A particle of mass `m` is projected from the ground with initial linear momentum `p` (magnitude) such that to have maximum possible range, its minimum kinetic energy will be

Options

A. `(p^2)/(2m)`
B. `(p^2)/(4m)`
C. `(p^2)/m`
D. `(p^2)/(3m)`

Answer

To solve the problem, we need to determine the minimum kinetic energy of a particle projected from the ground with an initial linear momentum \( p \) such that it achieves the maximum possible range.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \( u \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity.

2. **Maximizing the Range**:
The range is maximized when \( \sin 2\theta = 1 \). This occurs when \( 2\theta = 90^\circ \) or \( \theta = 45^\circ \).

3. **Relating Momentum and Kinetic Energy**:
The initial linear momentum \( p \) is related to the initial velocity \( u \) by:
\[ p = mu \]
Therefore, the initial velocity can be expressed as:
\[ u = \frac{p}{m} \]

4. **Calculating Kinetic Energy**:
The kinetic energy \( K \) of the particle is given by:
\[ K = \frac{1}{2} mu^2 \]
Substituting \( u \) from the momentum equation:
\[ K = \frac{1}{2} m \left(\frac{p}{m}\right)^2 = \frac{p^2}{2m} \]

5. **Kinetic Energy at Maximum Range**:
At the maximum range, the velocity components at the highest point of the trajectory are:
- Horizontal component: \( u \cos \theta \)
- Vertical component: \( u \sin \theta \) (which becomes 0 at the highest point)

Since \( \theta = 45^\circ \):
\[ u \cos 45^\circ = u \sin 45^\circ = \frac{u}{\sqrt{2}} \]
The kinetic energy at the highest point is:
\[ K_{\text{highest}} = \frac{1}{2} m \left(u \cos 45^\circ\right)^2 = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{2} = \frac{mu^2}{4} \]

6. **Substituting for \( u^2 \)**:
From the earlier kinetic energy expression:
\[ K_{\text{highest}} = \frac{m}{4} \cdot \left(\frac{p^2}{m^2}\right) = \frac{p^2}{4m} \]

### Final Answer:
Thus, the minimum kinetic energy required for the particle to achieve the maximum possible range is:
\[ \boxed{\frac{p^2}{4m}} \]