STATEMENT-1: A body is projected from the ground with kinetic energy K at an angle of 60° with the horizontal. If air resistance is neglected, its K.E. When it is at the highest point of its trajectory will be K/4
because
STATEMENT-2: At the highest point of the trajectory, the direction of velocity and acceleration of the body are perpendicular to each other

Options

A. Statement-1 is True, Statement-2 is True, Statement -2 is a correct explanation for
Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement -2 is NOT a correct explanation for
Statement-1
C. Statement - 1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Understanding Kinetic Energy at Projection
The body is projected with an initial kinetic energy \( K \) at an angle of \( 60^\circ \) with the horizontal. The initial kinetic energy can be expressed as:
\[ K = \frac{1}{2} m v^2 \]
where \( v \) is the initial velocity of the body and \( m \) is its mass.

**Hint:** Remember that kinetic energy is related to the mass and the square of the velocity.

### Step 2: Finding Horizontal and Vertical Components of Velocity
The initial velocity \( v \) can be resolved into horizontal and vertical components:
- Horizontal component \( v_x = v \cos(60^\circ) = v \cdot \frac{1}{2} = \frac{v}{2} \)
- Vertical component \( v_y = v \sin(60^\circ) = v \cdot \frac{\sqrt{3}}{2} \)

**Hint:** Use trigonometric identities to resolve the velocity into components.

### Step 3: Velocity at the Highest Point
At the highest point of the trajectory, the vertical component of the velocity becomes zero (\( v_y = 0 \)), and only the horizontal component remains:
\[ v_{x, \text{highest}} = \frac{v}{2} \]

**Hint:** Remember that at the highest point, the object stops moving upward, so its vertical velocity is zero.

### Step 4: Kinetic Energy at the Highest Point
The kinetic energy at the highest point \( K_h \) can be calculated using the remaining horizontal velocity:
\[ K_h = \frac{1}{2} m v_{x, \text{highest}}^2 = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \cdot \frac{v^2}{4} = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{K}{4} \]

**Hint:** Substitute the expression for \( v_{x, \text{highest}} \) into the kinetic energy formula.

### Step 5: Analyzing Statement 1
Statement 1 claims that the kinetic energy at the highest point is \( \frac{K}{4} \). From our calculations, we found that this is indeed true.

**Hint:** Verify the calculations to ensure that the kinetic energy at the highest point matches the assertion.

### Step 6: Analyzing Statement 2
Statement 2 states that at the highest point of the trajectory, the direction of velocity and acceleration of the body are perpendicular to each other. At this point:
- The velocity is horizontal (along the x-axis).
- The acceleration due to gravity acts downward (along the y-axis).

Since the horizontal and vertical directions are perpendicular, this statement is also true.

**Hint:** Consider the directions of velocity and acceleration vectors to determine their relationship.

### Conclusion
Both statements are true:
- Statement 1 is true because the kinetic energy at the highest point is \( \frac{K}{4} \).
- Statement 2 is true because the velocity and acceleration are perpendicular at the highest point.

However, Statement 2 does not provide a correct explanation for Statement 1, as the relationship between the two is not direct.

**Final Answer:** Statement 1 is true, Statement 2 is true, but Statement 2 is not the correct explanation for Statement 1.