Answer
$F_{water}=2448N$
Work Step by Step
The person's dive can be divided into 2 periods:
1) Before she hits the water
This period does not have nonconservative forces, so energies are conserved: $$E_f-E_0=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2-v_0^2)+g(h_f-h_0)=0$$
We have $v_0=0$, $h_f-h_0=-3m$, and we need to find $v_f$ $$\frac{1}{2}v_f^2+9.8(-3)=0$$ $$v_f=7.67m/s$$
2) After she hits the water
In this period, the energies are not conserved. According to the work-energy theorem, $$E_f-E_0=W_{nc}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=W_{water}$$
Initial speed is the speed when the person hits the water at $v_0=7.67m/s$; $v_f=0$, $h_f-h_0=-1.1m$ and $m=67kg$. Therefore, $$W_{water}=-2693.03J$$ $$-F_{water}(h_f-h_0)=-2693.03J$$ $$F_{water}=\frac{2693.03}{1.1}=2448N$$
