A particle is projected making angle 45° with horizontal having kinetic energy K. The kinetic energy at highest point will be

The correct option is K/2.

Let us break down the solution step-by-step:

1. Initial Kinetic Energy:
Let $m$ be the mass of the particle and $u$ be its initial velocity of projection.
The initial kinetic energy $K$ of the particle is given by:
$K=\frac{1}{2}m{u}^2$

2. Velocity at the Highest Point:
When a particle is projected at an angle $\theta$ with the horizontal, it undergoes projectile motion.
Throughout the motion, the horizontal component of velocity remains constant because there is no horizontal acceleration (neglecting air resistance).
The horizontal component of the initial velocity is:
$u_x=u\cos \left(\theta \right)$
At the highest point of its trajectory, the vertical component of velocity becomes zero ($v_y=0$).
Therefore, the velocity of the particle at the highest point is purely horizontal:
$v=u_x=u\cos \left(\theta \right)$

3. Kinetic Energy at the Highest Point:
Let $K_{\text{top}}$ be the kinetic energy at the highest point.
$K_{\text{top}}=\frac{1}{2}m{v}^2=\frac{1}{2}m{\left(u\cos \theta \right)}^2$
Simplifying this, we get:
$K_{\text{top}}=\left(\frac{1}{2}m{u}^2\right){\cos }^2\left(\theta \right)$
Substituting $K=\frac{1}{2}m{u}^2$:
$K_{\text{top}}=K{\cos }^2\left(\theta \right)$

4. Calculation for the Given Angle:
We are given that the angle of projection is $\theta =45°$.
Since $\cos (45°)=\frac{1}{\sqrt{2}}$, we have:
${\cos }^2(45°)={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$
Substituting this value back into the expression for kinetic energy:
$K_{\text{top}}=K\cdot \frac{1}{2}=\frac{K}{2}$

Thus, the kinetic energy of the particle at the highest point of its trajectory is indeed $K/2$.