A projectile is projected with a kinetic energy K. Its range is R. It will have the minimum kinetic energy, after covering a horizontal distance equal to

The correct option is 0.5R.

When a projectile is launched, its velocity at any point along its trajectory can be resolved into a horizontal component and a vertical component. The horizontal component of the velocity, $v_x=ucos\theta$, remains constant throughout the motion because there is no horizontal force or acceleration acting on the projectile (neglecting air resistance).

The vertical component of the velocity, $v_y$, varies under the constant downward acceleration due to gravity. The kinetic energy of the projectile at any instant is determined by its mass and its speed, which depends on both velocity components:
$\mathrm{K.E.}=\frac{1}{2}m{v}^2=\frac{1}{2}m(v_x^2+v_y^2)$

Since the horizontal term $v_x^2$ is constant, the kinetic energy is at its minimum when the vertical velocity term $v_y^2$ is as small as possible. The minimum possible value for $v_y^2$ is zero, which occurs at the highest point of the flight where the vertical velocity component momentarily becomes zero ($v_y=0$).

Due to the parabolic symmetry of the trajectory, the projectile reaches its maximum height at exactly half of the total horizontal range ($R$). Thus, the horizontal distance covered when the projectile has its minimum kinetic energy is:
$\mathrm{Horizontal\; Distance}=\frac{R}{2}=0.5R$