## Physics (10th Edition)

The minimum distance is $170.73m$
The closest case scenario is that the woman makes her circular turn and finishes it right beside the seawall. In this case, the distance from the seawall that she begins making the turn equals the radius $r$ of her circular path. In other words, this exercise asks for this value of $r$. We have $v=26m/s$, $\theta=22^o$ and $g=9.8m/s^2$. We can find $r$ by $$\tan\theta=\frac{v^2}{rg}$$ $$r=\frac{v^2}{g\tan\theta}=170.73m$$