Answer: 6. Solution. From the motion law for a body thrown from ground level at an angle $\alpha$ to the horizontal, the range of flight is determined by the relation $L=\frac{V_{0}^{2}}{g} \sin 2 \alpha$. Therefore, the maximum range of flight is achieved at $\alpha=45^{\circ}$ and is equal to $L=\frac{V_{0}^{2}}{g}$. This means $V_{0}=\sqrt{g L}$.

The fragment that falls to the ground first is the one that flies vertically downward, and the last one is the one that flies vertically upward. After time $\tau=\frac{2 V_{0}}{g}$, the last fragment will return to the explosion site, which means it will be in the same situation as the first fragment. Therefore, the required time interval is $\tau=\frac{2 V_{0}}{g}$ (independent of height $H$). Thus, $\tau=\frac{2 \sqrt{g L}}{g}=2 \sqrt{\frac{L}{g}}=6$ seconds.
5.1. A grenade lying on the ground explodes into a multitude of small identical fragments, which scatter in a radius of $L=90$ m. Determine the time interval (in seconds) between the moments of impact on the ground of the first and the last fragment, if such a grenade explodes in the air at a height of $H=10 \mathrm{m}$. The acceleration due to gravity is considered to be $g=10$ m/s ${ }^{2}$. Air resistance is negligible.