Answer
$t_{min}=8.73s$
Work Step by Step
1) Find the acceleration $a$
The person is pulled up by tension force $T$, which is opposed by the person's weight $mg$. As the movement has acceleration $a$, we have $$T-mg=ma$$
As $T_{max}=569N$, $mg=520N$ and $m=\frac{520}{9.8}=53kg$, we find $a_{max}$ $$a_{max}=\frac{T_{max}-mg}{m}=0.92m/s^2$$
2) Find $t_{min}$
We have $a_{max}=0.92m/s^2$, $v_0=0$ and cave's depth $y=35.1m$
$$y=v_0t+1/2at^2=1/2at^2$$ $$t_{min}=8.73s$$
