A shot is fired at `30^@` with the vertical from a point on the ground with kinetic energy `K`. If air resistance is ignored, the kinetic energy at the top of the trajectory is

Options

A. `3K//4`
B. `K//2`
C. `K`
D. `K//4`

Explanation

To solve the problem of finding the kinetic energy at the top of the trajectory of a projectile fired at an angle of \(30^\circ\) with respect to the vertical, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial kinetic energy \(K\) of the projectile is given. The projectile is fired at an angle of \(30^\circ\) with the vertical. This means the angle with the horizontal is \(60^\circ\) (since \(90^\circ - 30^\circ = 60^\circ\)).

### Step 2: Break Down the Initial Velocity
Let the initial velocity of the projectile be \(u\). We can break this velocity into its horizontal and vertical components:
- The horizontal component \(u_x = u \cos(60^\circ) = u \cdot \frac{1}{2} = \frac{u}{2}\)
- The vertical component \(u_y = u \sin(60^\circ) = u \cdot \frac{\sqrt{3}}{2}\)

### Step 3: Analyze the Kinetic Energy at the Top of the Trajectory
At the top of the trajectory, the vertical component of the velocity becomes zero because the projectile momentarily stops moving upward before starting to descend. Therefore, only the horizontal component contributes to the kinetic energy at the top.

The kinetic energy at the top of the trajectory \(K'\) can be expressed as:
\[ K' = \frac{1}{2} m (u_x)^2 \]
Substituting the horizontal component:
\[ K' = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{4} = \frac{1}{8} m u^2 \]

### Step 4: Relate the Initial Kinetic Energy to the Kinetic Energy at the Top
The initial kinetic energy \(K\) is given by:
\[ K = \frac{1}{2} m u^2 \]

### Step 5: Find the Ratio of Kinetic Energies
To find the relationship between \(K\) and \(K'\):
\[ K' = \frac{1}{8} m u^2 = \frac{1}{4} \left(\frac{1}{2} m u^2\right) = \frac{K}{4} \]

### Conclusion
Thus, the kinetic energy at the top of the trajectory is:
\[ K' = \frac{K}{4} \]