1. Resolve the initial velocity

The projectile is thrown at an angle $45^\circ$ from the vertical with speed $5\sqrt{2}\,\text{m/s}$.

So the angle with the horizontal is also $45^\circ$.

Hence, $u_x = 5\sqrt{2}\cos45^\circ = 5\,\text{m/s}$ $u_y = 5\sqrt{2}\sin45^\circ = 5\,\text{m/s}$

2. Find the highest point

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

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

At the highest point, the vertical velocity is zero, and horizontal velocity remains $v_x = 5\,\text{m/s}$

3. Use the information after splitting

The projectile splits into two equal parts at the highest point.

One part falls vertically down and reaches the ground in $0.5\,\text{s}$.

Since it falls vertically, its horizontal velocity after splitting is zero.

Let mass of each part be $m$. Just before splitting, total horizontal momentum is $P_x = (2m)(5) = 10m$

After splitting, one part has horizontal velocity $0$. Let the horizontal velocity of the other part be $v$.

By conservation of horizontal momentum: $2m\cdot 5 = m\cdot 0 + m\cdot v$ $10m = mv$ $v = 10\,\text{m/s}$

4. Vertical motion after splitting

Before splitting, the vertical velocity is zero. The part that falls vertically reaches ground in $0.5\,\text{s}$ from height $1.25\,\text{m}$.

Let its initial vertical velocity just after splitting be $u_1$. Using $s = u_1 t + \frac{1}{2}gt^2$ with downward displacement $1.25\,\text{m}$ in $0.5\,\text{s}$: $1.25 = u_1(0.5) + \frac{1}{2}(10)(0.5)^2$ $1.25 = 0.5u_1 + 1.25$ $u_1 = 0$

So this part had zero vertical velocity just after splitting.

Since vertical momentum is also conserved and initial vertical momentum at the top is zero, $m\cdot 0 + m\cdot u_2 = 0$ which gives $u_2 = 0$

Thus the second part also starts with zero vertical velocity from the same height $1.25\,\text{m}$.

5. Time taken by the second part to hit the ground

For the second part, vertical motion is simply free fall from height $1.25\,\text{m}$ with zero initial vertical velocity: $1.25 = \frac{1}{2}gt^2 = 5t^2$ $t^2 = \frac{1.25}{5} = 0.25$ $t = 0.5\,\text{s}$

Therefore, $\boxed{t=0.5\,\text{s}}$