Understanding the Problem
To solve this problem, we need to analyze the motion of the bullet as it penetrates the target. The bullet loses half of its velocity after penetrating 6 cm, indicating that it experiences constant resistance.

Initial Conditions
- Let the initial velocity of the bullet be \( v_0 \).
- After 6 cm, the velocity becomes \( v_1 = \frac{1}{2} v_0 \).

Constant Resistance and Equation of Motion
The bullet faces a constant resistance force while moving through the target. This can be modeled by the equation of motion based on the work-energy principle:
- The work done against resistance is equal to the change in kinetic energy.
Using the formula for kinetic energy:
- \( KE = \frac{1}{2} mv^2 \)
The change in kinetic energy when the bullet slows from \( v_0 \) to \( v_1 \) over the first 6 cm is:
- \( \Delta KE = \frac{1}{2} m v_0^2 - \frac{1}{2} m \left(\frac{1}{2} v_0\right)^2 \)
- \( = \frac{1}{2} m v_0^2 - \frac{1}{8} m v_0^2 \)
- \( = \frac{3}{8} m v_0^2 \)

Further Penetration Calculation
The bullet will continue to penetrate until it comes to rest (final velocity \( v = 0 \)). The change in kinetic energy from \( v_1 \) to rest is:
- \( \Delta KE = \frac{1}{2} m \left(\frac{1}{2} v_0\right)^2 - 0 \)
- \( = \frac{1}{8} m v_0^2 \)
Since the resistance is constant, the distance \( d \) for this penetration can be calculated using the ratio of the kinetic energy changes:
- Let \( d \) be the additional penetration distance.
- The ratio of the distances can be derived from the energies:
Using the relationship:
- \( \frac{d_1}{d_2} = \frac{\Delta KE_1}{\Delta KE_2} \)
Where:
- \( d_1 = 6 \, \text{cm} \) and \( \Delta KE_1 = \frac{3}{8} m v_0^2 \)
- \( d_2 = d \) and \( \Delta KE_2 = \frac{1}{8} m v_0^2 \)
Thus, we get:
- \( \frac{6}{d} = \frac{\frac{3}{8}}{\frac{1}{8}} \)
- \( \frac{6}{d} = 3 \)
This leads to:
- \( d = \frac{6}{3} = 2 \, \text{cm} \)

Conclusion
The bullet will penetrate an additional **2 cm** before coming to rest.