You have approximately 29 tons of fish on deck. What will be the shift in the center of gravity after you shift the fish to the fish hold, a vertical distance of 5 feet? (total displacement is 483 tons)

Correct Answer: Option C — 0.3 foot

Applying the vertical center of gravity formula $\Delta KG = (w \times d) / W$, multiply 29 tons by the 5-foot vertical distance and divide by the 483 tons total displacement, resulting in a 0.3 foot shift.