Mass-spring-damper problem for kayaking waterfalls

[Baluncore]

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[jack action]

However, I suspect the water doesn't act like a spring but more like a damper with uniform rate of dampening during the entire impact.

OK, let's model the water as a pure damper, that is $F= -Cv$ where $C$ is a damping coefficient.

We already know the initial velocity upon impact, i.e. $v_0 = \sqrt{2gh}$. Knowing that $v = \frac{dx}{dt}$ and $F = m\frac{dv}{dt}$, then:
$$F = mv\frac{dv}{dx}$$
$$-Cv = mv\frac{dv}{dx}$$
$$dx = -\frac{m}{C}dv$$
$$\int_0^d dx = -\frac{m}{C}\int_{v_0}^0 dv$$
$$d = \frac{m}{C}v_0$$
Thus we can estimate $C$ if we know the distance $d$ that the kayak sinks in:
$$C = \frac{mv_0}{d}$$
The maximum damping force will be at maximum velocity, which is at $v_0$:
$$F_{d\ max} = C v_0 = \frac{m}{d}v_0^2 = \frac{2mgh}{d} = \frac{2Wh}{d}$$
For $W= 175\ lb$, $h = 20\ ft$ and $d= 6\ in$, then $F_{d\ max} = 14\ 000\ lb$.

What difference will adding a spring in series do? Assuming the ideal case where the spring compresses completely before the kayak begins sinking:
$$v_0 = \sqrt{2gh - \frac{K_s}{m}x_s^2}$$
and
$$F_{s\ max}= K_s x_s$$
$$F_{d\ max} = \frac{m\sqrt{2gh}}{d}\sqrt{2gh - \frac{K_s}{m}x_s^2}$$
The spring constant necessary to achieve a compression of $2"$ without bottoming out is $5930\ lb$ which will give $F_{s\ max} = F_{d\ max} = 11\ 860\ lb$, same as with the 2 springs in series. With any other spring constant, either $F_{s\ max}$ or $F_{d\ max}$ will increase (up to $14\ 000\ lb$).

I thought that was an interesting result. I guess it means that there is a maximum value for the energy that you can absorb with a spring under the seat, no matter how the energy will be dissipated afterward.