We model the salmon's leap as a projectile motion problem under constant gravitational acceleration. The key is to find the maximum vertical displacement.

[Q1] Maximum height of the waterfall We use the time-independent kinematic equation relating initial velocity ($v_0$), final velocity ($v_f$), acceleration ($a$), and displacement ($\Delta y$):

$$v_f^2 = v_0^2 + 2a\Delta y$$

At the maximum height, $h_{max}$, the salmon's vertical velocity is momentarily zero, so $v_f = 0$. The acceleration is due to gravity, pointing downwards, so $a = -g$. The displacement is the maximum height, $\Delta y = h_{max}$.

Substituting these into the equation:

$$0^2 = v_0^2 + 2(-g)h_{max}$$ $$v_0^2 = 2gh_{max}$$

Solving for the maximum height, we derive the expression:

$$h_{max} = \frac{v_0^2}{2g}$$

Next, we convert the initial speed from km/h to m/s:

$$v_0 = 32 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = \frac{32 \times 1000}{3600} \text{ m/s} = \frac{80}{9} \text{ m/s}$$

Now, we substitute the numerical values into our derived expression:

$$h_{max} = \frac{\left(\frac{80}{9} \text{ m/s}\right)^2}{2(9.8 \text{ m/s}^2)} = \frac{\frac{6400}{81} \text{ m}^2/\text{s}^2}{19.6 \text{ m/s}^2} \approx 4.03 \text{ m}$$

[Q2] Comparison with human high jump record The human high jump record is $h_{human} = 2.45$ m. To compare the salmon's leap to the human record, we can find the ratio:

$$\text{Ratio} = \frac{h_{max}}{h_{human}} = \frac{4.03 \text{ m}}{2.45 \text{ m}} \approx 1.64$$

The salmon can leap approximately 1.64 times higher than the men's human high jump record.