smc

Assume that the two boats are traveling along the positive real number line, with the merchantman starting at the number $10$ and the privateer starting at the number $0$. After two hours the merchantman is at $26$ while the privateer is at $22$. When the top sail of the privateer is carried away, the speed of the merchantman is unaffected; he still travels at $8$ miles per hour. Therefore the privateer travels at $17\cdot \frac{8}{15}=\frac{136}{15}=9\frac{1}{15}$ miles per hour. The remaining time $t$ in hours it takes for the privateer to catch up to the merchantman satisfied the equation $$26+8t=22+\frac{136}{15}t$$ Simplification yields the equation $\frac{16}{15}t=4$, which shows that $t=\frac{15}{4}=3\frac{3}{4}$. It therefore takes a total of $5\frac{3}{4}$ hours for the privateer to catch the merchantman, so this will happen at $\boxed{5\text{:}30\text{ p.m.}}$