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, we need to find the minimum kinetic energy of an object projected at an angle such that its maximum height is one-fourth of its horizontal range. Let's break down the solution step by step.

### Step 1: Understand the relationship between maximum height and range
Given that the maximum height (H) is one-fourth of the horizontal range (R), we can express this relationship mathematically:
\[ H = \frac{1}{4} R \]

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

### Step 3: Substitute the relationship into the equations
From the relationship \( H = \frac{1}{4} R \), we can substitute the formulas:
\[ \frac{u^2 \sin^2 \theta}{2g} = \frac{1}{4} \left( \frac{u^2 \sin 2\theta}{g} \right) \]

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

Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \sin^2 \theta = \frac{1}{4} (2 \sin \theta \cos \theta) \]
\[ \sin^2 \theta = \frac{1}{2} \sin \theta \cos \theta \]

### Step 5: Rearranging the equation
Dividing both sides by \( \sin \theta \) (assuming \( \sin \theta \neq 0 \)):
\[ \sin \theta = \frac{1}{2} \cos \theta \]
\[ \tan \theta = \frac{1}{2} \]

### Step 6: Find the angle of projection
From \( \tan \theta = \frac{1}{2} \), we can find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]

### Step 7: Calculate the velocity at the highest point
At the highest point of the projectile, the vertical component of the velocity is zero. The horizontal component remains:
\[ v_x = u \cos \theta \]

### Step 8: Find the kinetic energy at the highest point
The kinetic energy (KE) at the highest point is given by:
\[ KE = \frac{1}{2} m v^2 \]
Substituting for \( v \):
\[ KE = \frac{1}{2} m (u \cos \theta)^2 \]
\[ KE = \frac{1}{2} m u^2 \cos^2 \theta \]

### Step 9: Express \( u^2 \) in terms of momentum \( p \)
The momentum \( p \) is given by:
\[ p = m u \]
Thus,
\[ u = \frac{p}{m} \]
Substituting this into the kinetic energy equation:
\[ KE = \frac{1}{2} m \left(\frac{p}{m}\right)^2 \cos^2 \theta \]
\[ KE = \frac{p^2 \cos^2 \theta}{2m} \]

### Step 10: Find \( \cos^2 \theta \)
Using \( \tan \theta = \frac{1}{2} \):
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Let \( \sin \theta = \frac{1}{\sqrt{5}} \) and \( \cos \theta = \frac{2}{\sqrt{5}} \):
\[ \cos^2 \theta = \frac{4}{5} \]

### Step 11: Substitute \( \cos^2 \theta \) back into the kinetic energy equation
\[ KE = \frac{p^2 \cdot \frac{4}{5}}{2m} \]
\[ KE = \frac{2p^2}{5m} \]

### Final Answer
The minimum kinetic energy in the path of the projectile is:
\[ KE = \frac{2p^2}{5m} \]