A body projected at an angle `theta` to the horizontal with kinetic energy `E_k` . The potential energy at the highest point of the trajectory is

Options

A. `E_k`
B. `E_kcos^2 theta`
C. `E_k sin^2theta`
D. `E_ktan^2theta`

Answer

To find the potential energy at the highest point of the trajectory of a body projected at an angle \( \theta \) to the horizontal with kinetic energy \( E_k \), we can follow these steps:

### Step 1: Understand the Initial Kinetic Energy
The initial kinetic energy \( E_k \) of the body can be expressed as:
\[ E_k = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the body and \( v \) is its initial velocity.

### Step 2: Express Velocity in Terms of Kinetic Energy
From the expression for kinetic energy, we can solve for \( v^2 \):
\[ v^2 = \frac{2E_k}{m} \]

### Step 3: Resolve Velocity into Components
When the body is projected at an angle \( \theta \), its velocity can be resolved into horizontal and vertical components:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = v \sin \theta \)

### Step 4: Determine the Vertical Component of Velocity
At the highest point of the trajectory, the vertical component of the velocity becomes zero. Therefore, we can use the vertical component \( v_y \) to find the maximum height \( h \) reached.

### Step 5: Use the Third Equation of Motion
Using the third equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity at the highest point)
- \( u = v_y = v \sin \theta \) (initial vertical velocity)
- \( a = -g \) (acceleration due to gravity, acting downwards)
- \( s = h \) (height reached)

Substituting these values, we get:
\[ 0 = (v \sin \theta)^2 - 2gh \]
Rearranging gives:
\[ h = \frac{(v \sin \theta)^2}{2g} \]

### Step 6: Substitute for \( v^2 \)
Now, substitute \( v^2 = \frac{2E_k}{m} \) into the height equation:
\[ h = \frac{(v \sin \theta)^2}{2g} = \frac{(v^2 \sin^2 \theta)}{2g} = \frac{\left(\frac{2E_k}{m}\right) \sin^2 \theta}{2g} = \frac{E_k \sin^2 \theta}{mg} \]

### Step 7: Calculate Potential Energy at Height \( h \)
The potential energy \( PE \) at height \( h \) is given by:
\[ PE = mgh \]
Substituting our expression for \( h \):
\[ PE = mg \left(\frac{E_k \sin^2 \theta}{mg}\right) = E_k \sin^2 \theta \]

### Final Answer
Thus, the potential energy at the highest point of the trajectory is:
\[ PE = E_k \sin^2 \theta \]