1. Resolve the initial velocity

The projectile is thrown at $45^\circ$ from the vertical, so it is also at $45^\circ$ above the horizontal.

Given speed: $u = 5\sqrt{2}\ \text{m/s}$

Hence horizontal and vertical components are $u_x = u\cos 45^\circ = 5\sqrt{2}\cdot \frac{1}{\sqrt{2}} = 5\ \text{m/s}$ $u_y = u\sin 45^\circ = 5\sqrt{2}\cdot \frac{1}{\sqrt{2}} = 5\ \text{m/s}$

2. Find the highest point of the original projectile

Time to reach highest point: $t_h = \frac{u_y}{g} = \frac{5}{10} = 0.5\ \text{s}$

Horizontal distance of highest point from $O$: $x_h = u_x t_h = 5\times 0.5 = 2.5\ \text{m}$

Maximum height: $H = \frac{u_y^2}{2g} = \frac{25}{20} = 1.25\ \text{m}$

3. Use the information after splitting

At the highest point, the projectile splits into two equal parts.

One part falls vertically down to the ground in $0.5\,$s after splitting.

Since it falls vertically, its horizontal velocity after splitting is zero. Also, because the highest point has zero vertical velocity before splitting, and the piece falls vertically in time $\sqrt{\frac{2H}{g}} = \sqrt{\frac{2(1.25)}{10}} = \sqrt{0.25} = 0.5\ \text{s},$ this piece simply starts from rest at the top and drops vertically.

4. Apply conservation of momentum at splitting

Just before splitting, at the highest point:

• vertical velocity = $0$
• horizontal velocity = $5\,$m/s

Let total mass be $M$. After splitting, each part has mass $M/2$.

Let horizontal velocity of the second part be $v$. The first part has horizontal velocity $0$.

Conservation of horizontal momentum: $M\cdot 5 = \frac{M}{2}\cdot 0 + \frac{M}{2}\cdot v$ $5 = \frac{v}{2}$ $v = 10\ \text{m/s}$

Vertical momentum before splitting is zero, and the first part also has zero vertical velocity, so the second part must also have zero vertical velocity immediately after splitting.

Thus the second part starts from the highest point with:

• horizontal velocity $10\,$m/s
• vertical velocity $0$

5. Time taken by the second part to fall

It starts from height $H=1.25\,$m with zero vertical velocity, so $H = \frac{1}{2}gt^2$ $1.25 = 5t^2$ $t^2 = 0.25$ $t = 0.5\ \text{s}$

6. Horizontal distance travelled by the second part after splitting

$\Delta x = vt = 10\times 0.5 = 5\ \text{m}$

So its final distance from point $O$ is $x = x_h + \Delta x = 2.5 + 5 = 7.5\ \text{m}$

Therefore, $\boxed{x=7.5}$