## Essential University Physics: Volume 1 (3rd Edition)

We know that $t=\sqrt{\frac{2h}{g}}$ The horizontal distance covered by the projectile is given as $x=v_{\circ}t$ $\implies x=v_{\circ}\sqrt{\frac{2h}{g}}$ $x=\sqrt{\frac{2v_{\circ}^2 h}{g}}$