1. **Problem Statement:** The speed of a projectile at the highest point is given as $\frac{1}{\sqrt{2}}$ times its initial speed $u$. We need to find the horizontal range of the projectile. 2. **Key Concepts and Formulas:** - Initial speed: $u$ - At the highest point, vertical velocity component is zero, so speed equals horizontal component. - Horizontal velocity $v_x = u \cos \theta$ - Vertical velocity $v_y = u \sin \theta$ - Speed at highest point $= v_x = u \cos \theta$ - Given speed at highest point $= \frac{u}{\sqrt{2}}$ - Horizontal range $R = \frac{u^2 \sin 2\theta}{g}$ 3. **Find the angle $\theta$:** Since speed at highest point is horizontal velocity, $$u \cos \theta = \frac{u}{\sqrt{2}}$$ Divide both sides by $u$: $$\cos \theta = \frac{1}{\sqrt{2}}$$ This implies: $$\theta = 45^\circ$$ 4. **Calculate horizontal range $R$:** Using formula: $$R = \frac{u^2 \sin 2\theta}{g}$$ Since $\theta = 45^\circ$, then $2\theta = 90^\circ$ and $\sin 90^\circ = 1$: $$R = \frac{u^2 \times 1}{g} = \frac{u^2}{g}$$ 5. **Answer:** The horizontal range of the projectile is $\boxed{\frac{u^2}{g}}$ which corresponds to option A.