The range of a projectile is 100   m . Its kinetic energy will be maximum after covering a distance of:

The expression for a horizontal range of a projectile is, $R=\frac{u^2\sin \left(2\theta \right)}{g}$here, $\theta$is the angle of projection and $u$is the initial resultant velocity.

The expression for kinetic energy is, $KE=\frac{1}{2}m{v}^2$

For the maximum kinetic energy, the velocity should be maximum, now the velocity of the projectile is $u=\sqrt{u_x^2+u_y^2}$

In the expression of the velocity of particle during projectile, the horizontal component of velocity is constant, i.e., $u_x=\text{constant}$therefore, net velocity depends upon the vertical component of velocity, i.e., $u^2\propto u_y^2\text{ or} u^2\cos \theta$

Therefore, when $u_y$is maximum then the net velocity will be maximum and so is the kinetic energy.

Since, $u_y$is maximum at the beginning and at the end of the flight of projectile, at the end of the flight of projectile the particle has already covered its horizontal range, therefore, kinetic energy will be maximum after covering a distance of $100 \mathrm{m}.$