The value of 𝑡 is ___.

Let the projectile be projected from the ground with speed $$u = 5\sqrt{2}\,\text{m s}^{-1}$$ at an angle of $$45^{\circ}$$ from the vertical (that is, $$45^{\circ}$$ from the horizontal as well).
Initial components: $$u_x = u\sin 45^{\circ} = 5\,\text{m s}^{-1}, \quad u_y = u\cos 45^{\circ} = 5\,\text{m s}^{-1}$$.

Maximum-height point (splitting point)
Time taken to reach the highest point is $$t_1 = \dfrac{u_y}{g} = \dfrac{5}{10} = 0.5\ \text{s}$$.
Maximum height reached above the ground is $$H = \dfrac{u_y^{\,2}}{2g} = \dfrac{25}{20} = 1.25\ \text{m}$$.
At this point the projectile has only horizontal velocity $$v_{\text{cm}} = u_x = 5\ \text{m s}^{-1}$$.

The projectile now breaks into two equal parts of mass $$\dfrac{m}{2}$$ each.

Velocities just after splitting
Because the splitting force is internal, linear momentum is conserved at that instant.
If $$\mathbf{v}_1$$ and $$\mathbf{v}_2$$ are the velocities of the two fragments just after splitting, momentum conservation gives
$$\frac{m}{2}\mathbf{v}_1 + \frac{m}{2}\mathbf{v}_2 = m\,\mathbf{v}_{\text{cm}}.$$
⟹ $$\mathbf{v}_1 + \mathbf{v}_2 = 2\mathbf{v}_{\text{cm}} = 10\,\hat{\mathbf{i}}\ \text{m s}^{-1}$$ $$-(1)$$.

The first fragment falls vertically down and reaches the ground in $$0.5\ \text{s}$$. Taking upward as positive and applying $$s = v_0 t + \dfrac{1}{2}(-g)t^{2}$$ to this fragment:

Displacement $$s = -1.25\ \text{m},\quad t = 0.5\ \text{s}$$,
$$-1.25 = v_{1y}(0.5) - 5(0.5)^2 = 0.5v_{1y} - 1.25$$
⟹ $$v_{1y} = 0$$. It also has no horizontal component (it “falls vertically”), so $$\mathbf{v}_1 = 0$$.

Substituting $$\mathbf{v}_1 = 0$$ in (1): $$\mathbf{v}_2 = 10\,\hat{\mathbf{i}}\ \text{m s}^{-1}.$$ Therefore the second fragment is projected horizontally with speed $$10\ \text{m s}^{-1}$$ from height $$1.25\ \text{m}$$.

Time taken by the second fragment to reach the ground
For purely vertical motion of this fragment, initial vertical speed $$v_{2y}=0$$ and displacement $$-1.25\ \text{m}$$:
$$-1.25 = 0\cdot t + \frac{1}{2}(-10)t^{2}$$
$$\Rightarrow t^{2} = \frac{2(1.25)}{10} = 0.25$$
$$\Rightarrow t = 0.5\ \text{s}.$$

Thus the second fragment reaches the ground $$t = 0.5\ \text{s}$$ after the splitting. This lies in the required interval $$0.45 - 0.55$$.

Answer: $$0.5\ \text{s}.$$