The external forces (gravity) act identically on the shell before and after explosion. Thus, the CM follows the original parabolic trajectory.

Initial Conditions:

• Initial Height $y_0 = 15 \text{ m}$
• Horizontal Velocity $u_x = 165 \text{ m/s}$
• Vertical Velocity $u_y = 0 \text{ m/s}$
• Projection Point: $(0, 15, 0)$

Position of CM at $t = 1.5 \text{ s}$:

$$ X_{cm} = u_x t = 165 \times 1.5 = 247.5 \text{ m} $$

$$ Y_{cm} = y_0 – \frac{1}{2}gt^2 = 15 – \frac{1}{2}(10)(1.5)^2 = 15 – 11.25 = 3.75 \text{ m} $$

$$ Z_{cm} = 0 \text{ m} $$

So, $\vec{R}_{cm} = (247.5, 3.75, 0)$.

The shell splits into two identical fragments, so $m_1 = m_2 = m/2$.

From the definition of Center of Mass:

$$ \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} = \frac{\vec{r}_1 + \vec{r}_2}{2} $$

Rearranging for the unknown fragment position $\vec{r}_2$:

$$ \vec{r}_2 = 2\vec{R}_{cm} – \vec{r}_1 $$

Fragment 1 Position ($\vec{r}_1$):

• Distance from wall ($x$): 240 m
• On the ground ($y$): 0 m
• 7 m to the right of x-axis ($z$): 7 m (Assuming right is positive z)
• $\vec{r}_1 = (240, 0, 7)$

Applying the formula:

$$ x_2 = 2(247.5) – 240 = 495 – 240 = 255 \text{ m} $$ $$ y_2 = 2(3.75) – 0 = 7.5 \text{ m} $$ $$ z_2 = 2(0) – 7 = -7 \text{ m} $$