An object of mass m is projected with a momentum p at such an angle that its maximum height is `1//4` th of its horizontal range. Its minimum kinetic energy in its path will be

Options

A. `(p^(2))/(8m)`
B. `(p^(2))/(4m)`
C. `(3p^(2))/(4m)`
D. `(p^(2))/(m)`

Explanation

To solve the problem step by step, we need to analyze the motion of the projectile and use the given conditions to find the minimum kinetic energy in its path.

### Step 1: Understand the relationship between maximum height and range
We are given that the maximum height \( H_{\text{max}} \) is \( \frac{1}{4} \) of the horizontal range \( R \):
\[ H_{\text{max}} = \frac{1}{4} R \]

### Step 2: Write the formulas for maximum height and range
The formulas for maximum height and range for a projectile launched at an angle \( \theta \) with initial velocity \( u \) are:
\[ H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \]
\[ R = \frac{u^2 \sin 2\theta}{g} \]

### Step 3: Substitute the relationship into the formulas
Substituting \( H_{\text{max}} \) into the equation gives:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{4} \left( \frac{u^2 \sin 2\theta}{g} \right) \]
This simplifies to:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin 2\theta}{4g} \]

### Step 4: Simplify the equation
Cancelling \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin^2 \theta = \frac{1}{2} \sin 2\theta \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \sin^2 \theta = \frac{1}{2} (2 \sin \theta \cos \theta) \]
This simplifies to:
\[ \sin^2 \theta = \sin \theta \cos \theta \]

### Step 5: Factor the equation
Factoring gives:
\[ \sin \theta (\sin \theta - \cos \theta) = 0 \]
This implies either \( \sin \theta = 0 \) or \( \sin \theta = \cos \theta \). The second case gives:
\[ \tan \theta = 1 \implies \theta = 45^\circ \]

### Step 6: Find the minimum kinetic energy
The minimum kinetic energy occurs at the highest point of the projectile's path, where the vertical component of the velocity is zero. The horizontal component of the velocity at this point is:
\[ v_x = u \cos \theta \]
Thus, the momentum at this point is:
\[ P = m v_x = m (u \cos \theta) \]
At \( \theta = 45^\circ \):
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \implies P = m \left( u \cdot \frac{1}{\sqrt{2}} \right) = \frac{m u}{\sqrt{2}} \]

### Step 7: Relate momentum to kinetic energy
The kinetic energy is given by:
\[ KE = \frac{P^2}{2m} \]
Substituting for \( P \):
\[ KE = \frac{\left( \frac{m u}{\sqrt{2}} \right)^2}{2m} = \frac{m^2 u^2 / 2}{2m} = \frac{m u^2}{4} \]

### Step 8: Express \( u \) in terms of \( P \)
Since \( P = m u \), we have:
\[ u = \frac{P}{m} \]
Substituting this back into the kinetic energy expression:
\[ KE = \frac{m \left( \frac{P}{m} \right)^2}{4} = \frac{P^2}{4m} \]

### Conclusion
Thus, the minimum kinetic energy in the path of the projectile is:
\[ \boxed{\frac{P^2}{4m}} \]