Physics Solution: Projectile Explosion

Step-by-Step Analysis

1. The Original Trajectory
The shell is fired and follows a standard parabolic path. Let the total time of flight be $T$ and the horizontal range be $R$. The highest point is reached at $t = T/2$.

2. The Explosion
The explosion occurs at a time $t_0$ before reaching the highest point. Thus, $t_0 < T/2$. The shell breaks into two fragments, A and B. Since the explosion is an internal force, the Center of Mass (COM) continues on the original parabolic path.

3. Motion of Fragment A
Fragment A comes to instantaneous rest. This means its velocity immediately after the explosion is $\vec{v}_A = 0$. It then falls vertically under gravity from height $h$ (the height at explosion).

4. Comparing Time Scales
We need to determine where the COM is when Fragment A hits the ground.

• Time remaining for the COM to complete the full trajectory: $t_{rem} = T – t_0$.
• Time for Fragment A to fall to the ground: $t_{fall}$.

Since the projectile was on the ascending part of the trajectory (before the highest point), the height $h$ is relatively low compared to the time remaining in the flight. Mathematically, for any point on the upward journey, the time it takes to drop from rest ($t_{fall}$) is less than the time it takes to complete the parabolic arc ($t_{rem}$). $$t_{fall} < t_{rem}$$

5. Conclusion
Fragment A hits the ground before the total time $T$ has elapsed.

• The COM reaches horizontal distance $R$ exactly at time $T$.
• Since the current time is $t = t_0 + t_{fall} < T$, the COM has not yet finished its journey.
• Therefore, the horizontal position of the COM, $x_{com}$, must be less than $R$.