1. **Problem statement:** We need to find the velocity with which a bullet strikes the ground, assuming it reaches terminal velocity before impact. 2. **Key concept:** Terminal velocity occurs when the acceleration is zero, meaning the forces on the bullet balance out and velocity becomes constant. 3. **Model assumption:** The bullet's velocity $v$ satisfies a differential equation of the form $$m \frac{dv}{dt} = mg - kv,$$ where $m$ is mass, $g$ is acceleration due to gravity, and $k$ is a drag coefficient. 4. **At terminal velocity $v_t$, acceleration is zero:** $$0 = mg - kv_t$$ 5. **Solve for terminal velocity:** $$kv_t = mg$$ $$v_t = \frac{mg}{k}$$ 6. **Interpretation:** The bullet strikes the ground with velocity $v_t = \frac{mg}{k}$ downward, so the velocity is negative if we take upward as positive. **Final answer:** $$v = -\frac{mg}{k}$$ (meters per second, downward)