Final answer:
The ball subjected to a buoyant force in water with negligible drag will rise but will not shoot higher than the depth it was released from. The rise is due to the work done by the buoyant force while submerged, and the maximum height achieved will be less than or equal to the depth of 0.730 m.
Step-by-step explanation:
The question asks how high above the water surface a ball will shoot as it emerges from the water, given that its density is 0.35 that of the water and there is negligible drag force acting on it.
To solve this, we use the conservation of mechanical energy, assuming energy is only exchanged between gravitational potential energy and buoyancy, and knowing that the buoyant force will do work on the ball to propel it upwards.
With the ball's density being 0.35 of water's, the buoyant force will exceed the gravitational force while the ball is submerged. However, the buoyancy vanishes as the ball emerges, and only the kinetic energy acquired underwater propels it upwards. The ball will not reach a height greater than the depth from which it was released due to losses when transitioning between mediums and the additional gravitational potential energy needed to be acquired above water.
So the answer is: the ball will not shoot above the water surface, and its maximum height will be less than or equal to 0.730 m.