Pessimist Singularitarian

#4
November 26th, 2016,
15:35
Yes:

$ \displaystyle s=r\theta=\left(40\text{ ft}\right)\left(2\cdot65^{\circ}\frac{\pi}{180^{\circ}}\right)=\frac{260\pi}{9}\,\text{ft}\approx90.76\text{ ft}$

To find the maximum height (above the lowest point) of the middle of the ship, we can observe that the height $h$ of the ship in terms of the angular displacement $\theta$ is given by:

$ \displaystyle h=r(1-\cos(\theta))$

So, use $r=40\text{ ft}$ and $\theta=65^{\circ}$ in the above formula to get the maximum height of the center of the ship.