Question

Suppose that you release a small ball from rest at a depth of 0.730 m below the surface in a pool of water. If the density of the ball is 0.350 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? (Neglect any transfer of energy to the splashing and waves produced by the emerging ball.)

Answer 1

Answer:

21.8 m

Explanation:

Net force on the ball inside water

= Buoyant force - weight of ball

= mρg - mdg

where ρ is density of water and d is density of ball

= m ( ρ - .35ρ ) g

This force acts on the ball upto distance .73 m so work done by force  will be equal to kinetic energy acquired by the ball.

m ( ρ - .35ρ ) g x .73 = 1/2 mv²

10³ ( 1 - .35 ) x g x .73 = 1/2 mv²

v² = 10³ x .65 x 9.8 x .73 x 2

v²  = 9300

After emerging from water , buoyant force stops acting on it.  

It will stop at height such that

1/2 mv² = mgh

v² = 2gh

h = v² / 2g

9300 / 2 x 9.8

= 21.8 m