vcsharp2003 said:
If we know that com will take the projectile path as if the explosion hadn't occurred, then can we not say with certainty that when com hits ground, then both exploded parts must also hit it otherwise the com would be above ground level due to the line joining the two parts being slanted to the ground?
Yes, the fact that the CoM hits the ground is no guarantee that the two fragments will hit the ground at the same time. Here is the mathematical proof that compares the unexploded trajectory with the trajectories of the to equal fragment.
Let ##V_{0x},~V_{0y}## be the components of the initial velocity. The maximum height where the explosion occurs is ##H=\dfrac{V_{0y}^2}{2g}##. The time of flight for the unexploded projectile is ##T_{\!f}=\dfrac{2V_{0y}}{g}## and the time elapsed until the explosion occurs is ##\frac{1 }{2}T_{\!f} = \dfrac{V_{0y}}{g}##.
Now assume that when the explosion breaks up the projectile into two equal pieces, piece A falls vertically to the ground it has initial velocity ##\vec v_A=\{0,-v_y\}## while piece B has velocity ##\vec v_B=\{2V_{0x},v_y\}## in accordance with momentum conservation. The vertical position of the CoM is $$Y_{\text{cm}}=\frac{\frac{m}{2}y_A+\frac{m}{2}y_B}{m}=\frac{1}{2}\left(H-v_yt-\frac{1}{2}gt^2\right)+\frac{1}{2}\left(H+v_yt-\frac{1}{2}gt^2\right).$$With ##H=\frac{V_{0y}^2}{2g}##, this becomes, ##Y_{\text{cm}}=\frac{V_{0y}^2}{2g}-\frac{1}{2}gt^2## which gives the time of flight for the CoM after the explosion, ##T_{\text{cm}}=\frac{V_{0y}}{g}=\frac{1}{2}T_{\!f}## as expected. The CoM will land horizontally at the same spot as the unexploded projectile as long as the fragments are in free fall for at least time ##\frac{1}{2}T_{\!f}.##
We can now find where the pieces are vertically when the CoM lands.
$$\begin{align} & y_A=\frac{V_{0y}^2}{2g}-v_y \frac{V_{0y}}{g}-\frac{1}{2}g\left(\frac{V_{0y}}{g}\right)^2=-v_y \frac{V_{0y}}{g}\nonumber \\ & y_B=\frac{V_{0y}^2}{2g}+v_y \frac{V_{0y}}{g}-\frac{1}{2}g\left(\frac{V_{0y}}{g}\right)^2=+v_y \frac{V_{0y}}{g}\nonumber \end{align}.$$This shows that if we know that the CoM hits the ground at the same spot as the unexploded projectile, then the two fragments will hit the ground at the same time
only if the explosion imparts no vertical momentum to them (##v_y=0##). It also shows that for the CoM to land at the same spot as the unexploded projectile when ##v_y\neq 0##, fragment A must fall through a hole in the ground as
@Steve4Physics has argued.
Now consider the hole-in-the-ground suggestion by
@Steve4Physics and assume that fragment A keeps falling through it until the fragment B lands. We know that the CoM lands at the same spot as the unexploded projectile at horizonttal distance ##\Delta X_{\text{cm}}=V_{0x}(\frac{1}{2}T_{\!f})=\dfrac{V_{0x}V_{0y}}{g}## from the point of the explosion. To find where fragment B lands, which is the goal of this problem, first we find its time of flight, $$0=\frac{V_{0y}^2}{2g}+v_y t_B-\frac{1}{2}gt_B^2\implies t_B=\frac{v_y+\sqrt{V_{0y}^2+v_y^2 }}{g}.$$Then we can find where the second fragment lands horizontally from the point of the explosion, $$\Delta X_B=2V_{0x}t_B=\frac{2V_{0x}\left(v_y+\sqrt{V_{0y}^2+v_y^2 }\right)}{g}.$$It should be clear that unless we know ##v_y##, we cannot predict where the second fragment will land. If we are not given ##v_y## explicitly we can infer that it is zero with statements such as
(a) Both fragments land at the same time;
(b) Fragment A drops straight down with zero initial vertical velocity;
(c) Fragment B has zero vertical initial velocity.
Lacking such statements, as is the case here, we have to assume that ##v_y=0##, otherwise we cannot answer the question.
Summary
This question can be answered directly by observing that fragment B takes the same amount amount of time to come down as it takes to go up to max height. Since its speed is doubled at max height, it must cover twice the horizontal distance during the down trip.
On edit: Corrected subscript typos and added the summary.
. However, if I were to pick nits, I would say that an object released from rest from a height is "falling vertically" throughout its motion until it stops. "Falling vertically" only establishes that the object has no instantaneous horizontal velocity component and says nothing about the initial vertical component. 
