# A 0.9 kg hollow ball with a radius of 0.12 m, filled with air, is released from rest at the bottom of a 2.0 m deep pool of water. How high above the water does the ball shoot upward? Neglect all frictional effects, and neglect the ball's motion when it is only partially submerged.

Neglect friction? Egads.

The average force on the ball is..

avg force=bouyancy at 1 m

= weightof water displaced - .9kg*g

= densitywater*4/3 PI r^3 - .9kg*g

Energy put in the ball= avg force(2m)

PE at top above water= .9g h solve for h.

## To find the height above the water to which the ball shoots upward, we need to calculate the average force on the ball and the energy put into the ball.

First, let's calculate the average force on the ball. The average force is equal to the buoyancy force at a depth of 1 meter. The buoyancy force is the weight of the water displaced by the ball. The weight of the water displaced is equal to the density of water times the volume of the ball times the acceleration due to gravity.

The volume of a hollow ball is given by the formula V = 4/3 * π * (R_outer^3 - R_inner^3), where R_outer is the radius of the outer surface of the ball and R_inner is the radius of the inner surface of the ball. In this case, since the ball is filled with air, the inner radius is 0.

So, the volume of the ball is V = 4/3 * π * R_outer^3.

Now, let's calculate the average force on the ball:

avg force = weight of water displaced - mass of the ball * acceleration due to gravity

= density of water * volume of the ball * acceleration due to gravity - mass of the ball * acceleration due to gravity

We can plug in the values:

density of water = 1000 kg/m^3 (approximate value)

R_outer = 0.12 m

mass of the ball = 0.9 kg

acceleration due to gravity = 9.8 m/s^2

avg force = 1000 kg/m^3 * (4/3 * π * 0.12^3) * 9.8 m/s^2 - 0.9 kg * 9.8 m/s^2

Now, let's calculate the energy put into the ball. The energy put into the ball is equal to the average force on the ball times the distance the ball travels, which is 2 meters in this case:

Energy put into the ball = avg force * distance

= avg force * 2 m

Finally, let's find the height above the water to which the ball shoots upward. The potential energy at the top (above water) is equal to the energy put into the ball. The potential energy is given by the formula PE = mass of the ball * acceleration due to gravity * height.

So, we can write:

PE at top above water = energy put into the ball

= mass of the ball * acceleration due to gravity * height

We can rearrange the equation to solve for the height:

Height = energy put into the ball / (mass of the ball * acceleration due to gravity)

Plug in the values:

Height = (avg force * 2 m) / (mass of the ball * acceleration due to gravity)

Now, substitute the value of avg force and solve for height.