A particle is projected from the ground with a kinetic energy E at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its motion will be

Options

A. `E//sqrt(2)`
B. `E//2`
C. `E//4`
D. `E//8`

Explanation

To find the kinetic energy of a particle projected from the ground with an initial kinetic energy \( E \) at an angle of \( 60^\circ \) with the horizontal, we can follow these steps:

### Step 1: Understand the Initial Conditions
The particle is projected with a kinetic energy \( E \). The total initial kinetic energy can be expressed in terms of the initial velocity \( u \):
\[ E = \frac{1}{2} m u^2 \]
where \( m \) is the mass of the particle.

### Step 2: Resolve the Initial Velocity
The initial velocity \( u \) can be resolved into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2} \)
- Vertical component: \( u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2} \)

### Step 3: Determine the Velocity at the Highest Point
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero (as the particle stops rising before it starts falling). Therefore, the velocity at the highest point is only the horizontal component:
\[ v_{top} = u_x = \frac{u}{2} \]

### Step 4: Calculate the Kinetic Energy at the Highest Point
The kinetic energy at the highest point can be calculated using the formula:
\[ K.E. = \frac{1}{2} m v_{top}^2 \]
Substituting \( v_{top} = \frac{u}{2} \):
\[ K.E. = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{4} = \frac{1}{8} m u^2 \]

### Step 5: Relate the Kinetic Energy at the Highest Point to \( E \)
Since we know \( E = \frac{1}{2} m u^2 \), we can express \( m u^2 \) in terms of \( E \):
\[ K.E. = \frac{1}{8} \cdot \frac{2E}{1} = \frac{E}{4} \]

### Conclusion
Thus, the kinetic energy at the highest point of its motion is:
\[ \boxed{\frac{E}{4}} \]