The speed of a projectile at the highest point becomes $\frac{1}{\sqrt{2}}$times its initial speed. The horizontal range of the projectile will be

The correct option is A

Velocity at the highest point is given by $ucos\theta =\frac{u}{\sqrt{2}}\left(given\right)\therefore \theta ={45}^{\circ }$

Horizontal range $R=\frac{u^{2}sin2\theta }{g}=\frac{u^{2}sin(2\times {45}^{\circ })}{g}=\frac{u^2}{g}$