The speed of a projectile at the highest point becomes 1/√2 times its initial speed. The horizontal range of the projectile will be

Correct Option: u²/g

Let the initial speed of the projectile be $u$ and the angle of projection with the horizontal be $\theta$.

The horizontal component of the velocity remains constant throughout the flight and is given by:
$v_x=u\cos \theta$
The vertical component of the velocity becomes zero at the highest point of the trajectory ($v_y=0$).

Therefore, the speed of the projectile at the highest point is purely horizontal:
$v=u\cos \theta$

According to the problem, the speed at the highest point is $\frac{1}{\sqrt{2}}$ times the initial speed:
$u\cos \theta =\frac{u}{\sqrt{2}}$

Simplifying the equation gives:
$\cos \theta =\frac{1}{\sqrt{2}}$
This implies that the angle of projection is:
$\theta =45°$

The horizontal range ($R$) of a projectile is given by the formula:
$R=\frac{u^2\sin \left(2\theta \right)}{g}$

Substituting $\theta =45°$ into the range formula:
$R=\frac{u^2\sin (2\times 45°)}{g}$

Since $\sin (90°)=1$, we have:
$R=\frac{u^2\times 1}{g}=\frac{u^2}{g}$

Thus, the horizontal range of the projectile is $\frac{u^2}{g}$.