#### Answer

308 m/s

#### Work Step by Step

From the conservation of momentum, we know that the initial momentum of the bullet and the pendulum combined must equal the combined momentum of the bullet embedded in the pendulum at the end.
Therefore we do the following:
$m_{bullet}v_{bullet} + m_{pendulum}v_{pendulum} = m_{both}v_{both}$
we know that $m_{bullet} = 10g = 0.01 kg$, $m_{pendulum} = 2kg$ and $m_{both} = 2.01 kg$.
Then we need to find the velocity of the system given that it rises 12cm vertically. 12cm can be converted to 0.12m and we know that the final velocity was 0 because it will follow a parabolic trajectory and come to a stop before falling back down and that the acceleration was -9.81 $m/s^2$ due to gravity.
Therefore, we can use
$v^2 = v_{0}^2 + 2a\Delta x$ to find the initial velocity and we find this to be 1.53m/s
Then we know that $m_{both}v_{both} = 3.08 N-s and that this equals the initial momentum so we have
3.08 N-s = 2kg*(0m/s) + 0.01v
and we find v to be 308 m/s