1. **Problem statement:** A bullet loses 75% of its velocity after penetrating 0.04 m into a wall. We need to find how far the bullet will penetrate the wall before coming to rest. 2. **Understanding the problem:** The bullet's velocity decreases as it penetrates the wall due to resistive forces. Losing 75% velocity means the bullet retains 25% of its initial velocity after 0.04 m. 3. **Assumptions and formula:** Assuming the resistive force is proportional to velocity, the velocity decreases exponentially with distance $x$: $$ v = v_0 e^{-kx} $$ where $v_0$ is initial velocity, $v$ is velocity after distance $x$, and $k$ is a positive constant. 4. **Using given data:** At $x=0.04$ m, velocity is 25% of $v_0$: $$ 0.25 v_0 = v_0 e^{-k \times 0.04} $$ Divide both sides by $v_0$: $$ 0.25 = e^{-0.04k} $$ Take natural logarithm: $$ \ln(0.25) = -0.04k $$ 5. **Calculate $k$:** $$ k = -\frac{\ln(0.25)}{0.04} $$ Since $\ln(0.25) = \ln(\frac{1}{4}) = -\ln(4)$, $$ k = -\frac{-\ln(4)}{0.04} = \frac{\ln(4)}{0.04} $$ 6. **Find total penetration distance $X$ where velocity becomes zero:** Velocity approaches zero as $x \to \infty$ in exponential decay, but physically bullet stops when velocity is zero. Alternatively, if resistive force is proportional to velocity squared or constant, penetration distance can be finite. Assuming velocity decreases linearly with distance for simplicity: Velocity lost per meter = $\frac{0.75 v_0}{0.04} = 18.75 v_0$ per meter. Total penetration distance $D$ when velocity reaches zero: $$ v_0 - 18.75 v_0 \times D = 0 $$ $$ 1 - 18.75 D = 0 $$ $$ D = \frac{1}{18.75} = 0.0533 \text{ m} $$ 7. **Answer:** The bullet will penetrate approximately 0.0533 meters (5.33 cm) into the wall before stopping. This is a simplified linear model based on given data.