Final answer:
The velocity of the sandal when it hits the deck, relative to the ship, is 17.15 m/s downward. Relative to a stationary observer on shore, it is approximately 17.20 m/s due south and downward after vector addition. Both perspectives agree on the point where the sandal hits the deck because of its consistent horizontal motion.
Step-by-step explanation:
The question involves calculating the velocity of a sandal when it hits the deck of a ship. To solve this, we consider the vertical and horizontal components separately. Since there is no horizontal force acting on the sandal, it retains the horizontal velocity of the ship, which is 1.75 m/s due south, relative to both the ship and the observer on shore. For the vertical component, we use the kinematic equation for free fall (ignoring air resistance): v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height (15.0 m).
Calculating this gives us v = √(2 * 9.8 m/s² * 15.0 m) = √(294 m²/s²) = 17.15 m/s downward. Hence, relative to the ship, the sandal's velocity when it hits the deck is 17.15 m/s downwards.
For part b), relative to a stationary observer on shore, we combine the downward velocity with the ship's horizontal motion using vector addition. We find the resultant velocity vector by calculating its magnitude using the Pythagorean theorem. The southward horizontal component remains 1.75 m/s, while the vertical component is 17.15 m/s downward. The magnitude of the resultant velocity is √(1.75² + 17.15²) m/s, which is approximately 17.20 m/s at an angle directed due south and downward.
Answer c) discusses consistency in results, explaining how both perspectives (ship and observer on shore) agree on the point where the sandal hits the deck because the horizontal motion of the sandal is the same as the ship's motion.