Solution

Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$.

According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop:
$$F \cdot h \approx mgH$$
$$F = mg\frac{H}{h}$$

Substituting the estimated values:
$$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$