At the apex of the trajectory, the shell's vertical velocity is zero, and its horizontal velocity is $v_x = v_0 \cos\theta$. Let the total mass be $2m$, so each fragment has mass $m$. Let the speeds of the fragments be $v_1$ and $v_2$. Momentum is conserved during the explosion.

Vertical momentum conservation ($p_y$): The initial vertical momentum is zero.

$$0 = m v_1 \sin(45^\circ) + m v_2 \sin(-45^\circ) = m \frac{\sqrt{2}}{2} (v_1 - v_2)$$

This implies $v_1 = v_2$. Let's call this speed $v'$.

Horizontal momentum conservation ($p_x$):

$$(2m)(v_0 \cos\theta) = m v_1 \cos(45^\circ) + m v_2 \cos(45^\circ)$$

Since $v_1=v_2=v'$,

$$2m v_0 \cos\theta = 2m v' \cos(45^\circ) = 2m v' \frac{\sqrt{2}}{2} = m v' \sqrt{2}$$

Solving for $v'$:

$$v' = \frac{2 v_0 \cos\theta}{\sqrt{2}} = \sqrt{2} v_0 \cos\theta$$