WW Kayaker said:
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.