A body is projected with kinetic energy E such that its range is maximum. Its potential energy at the maximum height is

Options

A. E
B. `E/2`
C. 2E
D. `sqrt(2)E`

Answer

To solve the problem, we need to find the potential energy of a body at its maximum height when projected with a kinetic energy \( E \) such that its range is maximum.

### Step-by-Step Solution:

1. **Understanding the Range of Projectile Motion**:
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 maximum when \( \sin 2\theta \) is maximum. The maximum value of \( \sin 2\theta \) is 1, which occurs when:
\[ 2\theta = 90^\circ \quad \Rightarrow \quad \theta = 45^\circ \]
Thus, the angle of projection for maximum range is \( 45^\circ \).

3. **Finding the Maximum Height**:
The maximum height \( h \) reached by the projectile is given by:
\[ h = \frac{U^2 \sin^2 \theta}{2g} \]
Substituting \( \theta = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \sin^2 45^\circ = \frac{1}{2} \]
Therefore, the maximum height becomes:
\[ h = \frac{U^2 \left(\frac{1}{2}\right)}{2g} = \frac{U^2}{4g} \]

4. **Calculating Potential Energy at Maximum Height**:
The potential energy \( PE \) at maximum height is given by:
\[ PE = mgh \]
Substituting for \( h \):
\[ PE = mg \left(\frac{U^2}{4g}\right) = \frac{mU^2}{4} \]

5. **Relating Potential Energy to Kinetic Energy**:
We know that the initial kinetic energy \( E \) is given by:
\[ E = \frac{1}{2} m U^2 \]
Rearranging gives:
\[ mU^2 = 2E \]
Substituting this into the potential energy equation:
\[ PE = \frac{2E}{4} = \frac{E}{2} \]

### Final Answer:
The potential energy at maximum height is:
\[ \text{Potential Energy} = \frac{E}{2} \]