Here is the mathematical treatment of this experiment. A block of mass $m_{block}$ is at rest when struck by a bullet of mass $m_{bullet}$ traveling with $v_{bullet}$ at a distance $x$ from the center of the block.
If the impact is perfectly plastic then the linear velocity of the block at the impact point should equal the velocity of the bullet after the impact. If the momentum transferred from the bullet to the block is $J$ then the change in speed for the bullet during the impact is
$$ \Delta v_{bullet} = - \frac{J}{m_{bullet}} $$
The same momentum $J$ affects the block by imparting linear and angular velocity at the center of gravity of
$$ \begin{aligned} \Delta v_{block} & = \frac{J}{m_{block}}
\\ \Delta \omega_{block} & = \frac{x\, J}{I_{block}} \end{aligned} $$
where $I_{block}$ is the mass moment of inertia of the block.
For a purely plastic contact setting the final relative speed equal to zero means
$$ \begin{aligned}
\Delta v_{block} + x \, \Delta \omega_{block} & = v_{bullet} + \Delta v_{bullet} \\
\frac{J}{m_{block}} + \frac{x^2\, J}{I_{block}} & = v_{bullet} - \frac{J}{m_{bullet}}
\end{aligned} $$
which is solved for:
$$ J = \frac{v_{bullet}}{ \frac{1}{m_{bullet}} + \frac{1}{m_{block}} + \frac{x^2}{I_{block}} } $$
Now for the trick. The combined center of gravity is at $x_{cg} = x \frac{ m_{bullet}} { m_{block} +m_{bullet} } $ so the take off speed of the new cg is
$$ \begin{aligned}
v_{cg} & = \Delta v_{block} + x_{cg} \Delta \omega_{block} \\
& = \frac{J}{m_{block}} + x \left( \frac{ m_{bullet}} { m_{block} +m_{bullet} } \right) \frac{x\, J}{I_{block}} \\
v_{cg} & =\left( \frac{ m_{bullet}} { m_{block} +m_{bullet} } \right) v_{bullet}
\end{aligned} $$
The above expression means that the take of speed (and hence the height) of the cg of the combined block and bullet does not depend on the impact location $x$.
:Pbut I just love your enthusiasm. All of you are awesome. I really admire physics and Physicists $\endgroup$ – Sourabh Aug 28 '13 at 16:55