The value of 𝑥 is ___.

The projectile is launched from point O with speed $$u = 5\sqrt{2}\text{ m s}^{-1}$$ at an angle $$45^{\circ}$$ with the vertical.
Hence the angle with the horizontal is also $$45^{\circ}$$.

Horizontal and vertical components of the initial velocity are therefore
$$u_x = u\cos 45^{\circ} = 5\sqrt{2}\cdot\frac{1}{\sqrt{2}} = 5\text{ m s}^{-1}$$
$$u_y = u\sin 45^{\circ} = 5\text{ m s}^{-1}$$

Time taken to reach the highest point (vertical velocity becomes zero):
$$t_h = \frac{u_y}{g} = \frac{5}{10} = 0.5\text{ s}$$

Height of the highest point above the ground:
$$H = \frac{u_y^2}{2g} = \frac{25}{20} = 1.25\text{ m}$$

At the highest point the projectile breaks into two equal parts (each of mass $$m$$).
External horizontal force is zero during the explosion, so horizontal momentum is conserved.

Horizontal momentum just before splitting:
$$p_{x,\text{before}} = (2m)u_x = 2m \times 5 = 10m$$

One fragment drops vertically (horizontal speed $$0$$).
Let the horizontal speed of the other fragment just after splitting be $$v_h$$. Conservation of horizontal momentum gives
$$10m = m(0) + mv_h \Longrightarrow v_h = 10\text{ m s}^{-1}$$

Vertical momentum before splitting is zero (vertical velocity was zero). With one fragment starting to fall from rest, the other fragment must also have zero initial vertical speed to keep the total vertical momentum zero. Thus both fragments start their downward fall from rest at the same height $$H$$.

Time taken by either fragment to reach the ground after splitting:
$$t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(1.25)}{10}} = \sqrt{0.25} = 0.5\text{ s}$$

This matches the given information for the first fragment and is also the fall time for the second fragment.

Horizontal travel of the projectile before splitting:
$$x_1 = u_x t_h = 5 \times 0.5 = 2.5\text{ m}$$

Horizontal travel of the second fragment after splitting:
$$x_2 = v_h t = 10 \times 0.5 = 5\text{ m}$$

Total horizontal distance of the second fragment from O:
$$x = x_1 + x_2 = 2.5 + 5 = 7.5\text{ m}$$

Hence the required distance lies in the range $$7.30 - 7.70\text{ m}$$.

Final Answer: $$x \approx 7.5\text{ m}$$ (within the given range).