Walk velocity

The units of the given quantities are $[v] = \frac{\text{m}}{\text{s}}, \quad [m] = \text{kg}, \quad [l] = \text{m}, \quad [g] = \frac{\text{m}}{\text{s}^2}$ It is noticeable that the unit kg appears only once in body weight. Therefore, the weight can not contribute to a dimensionless quantity and is irrelevant to solving the task. We obtain the dimensionless quantity from the other parameters $\alpha = \frac{v^2}{g l}, \quad [\alpha] = 1$ Solving the equation for the velocity results $v = \sqrt{\alpha g l} \propto l^{1/2} \quad \Rightarrow \quad \frac{v_2}{v_1} = \sqrt{\frac{l_2}{l_1}} \approx 1.0897$ Therefore, $v_2 \approx 1.0897 \cdot v_1 \approx 3 \,\frac{\text{m}}{\text{s}}$

During the walk, a part of the body acts like a pendulum, and the period is $T=\alpha \sqrt{l/g}$ , where $\alpha$ depends on the shape of the body, but not the size. The velocity is the ratio of the step size (proportional to $l$ ) to the period: $v=\alpha' \frac{l}{\sqrt{l/g}}=\alpha' \sqrt{lg}$ . Assuming $\alpha'$ is the same for Mr. and Mrs. Smith, we get $v_1/v_2=\sqrt{l_1/l_2}$ and $v_1=3.00$ m/s.

It is a bit confusing that Mr. Smith is so much shorter than Mrs. Smith.