Explain the principle of conservation of linear momentum and solve related problems.

Let the direction Hendrik is moving be positive.
Mass of Hendrik: $m_H = 65 \text{ kg}$
Mass of iron bar: $m_{bar} = 5 \text{ kg}$
Initial velocity of Hendrik and bar combined: $u = 1 \text{ m} \cdot \text{s}^{-1}$ (forward)
Total initial mass: $M = m_H + m_{bar} = 65 \text{ kg} + 5 \text{ kg} = 70 \text{ kg}$

To get the maximum decrease in his velocity, Hendrik throws the bar forward.
The speed at which he throws the bar away from himself is $4 \text{ m} \cdot \text{s}^{-1}$. This is the velocity of the bar relative to Hendrik.
Let Hendrik's final velocity be $v_H$.
The velocity of the bar relative to the ground, $v_{bar}$, is $v_H + 4 \text{ m} \cdot \text{s}^{-1}$.

The total momentum before throwing the bar equals the total momentum after throwing the bar.

$$M u = m_H v_H + m_{bar} v_{bar}$$ $$(70 \text{ kg})(1 \text{ m} \cdot \text{s}^{-1}) = (65 \text{ kg}) v_H + (5 \text{ kg}) (v_H + 4 \text{ m} \cdot \text{s}^{-1})$$ $$70 = 65 v_H + 5 v_H + 20$$ $$70 = 70 v_H + 20$$

$$70 v_H = 70 - 20$$ $$70 v_H = 50$$ $$v_H = \frac{50}{70} \text{ m} \cdot \text{s}^{-1}$$ $$v_H = \frac{5}{7} \text{ m} \cdot \text{s}^{-1}$$ $$v_H \approx 0.71 \text{ m} \cdot \text{s}^{-1}$$

Since $v_H$ is positive, Hendrik's final velocity is in the forward direction (the original direction of motion).

$$\boxed{\text{0.71 m}\cdot\text{s}^{-1} \text{ forward}}$$