How High Will the Ball Shoot Above the Water?

[fernancb]

Homework Statement

A 1.06 kg hollow ball with a radius of 0.113 m, filled with air, is released from rest at the bottom of a 1.94 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.

Homework Equations

P=pVg
H = total height of object
h= submerged height

The Attempt at a Solution

p(obj)Vg = p(H2O)Vg
175.4 kg/m^3 X A X H X g= 1000 kg/m^3 X A X h X g
h = 0.1754 H
= 0.0396 m

height above = 0.1864 m


Where am I going wrong?

 

[cepheid]

Here's a rho character that you can copy and paste: ρ

I'm a little confused by your formulae. Assuming that your small p was intended to be rho (density), then your equation for pressure (P) is incorrect. P ≠ ρVg. In fact, ρVg = mg, which has dimensions of force, not pressure. The correct expression is obtained by dividing this force by area:

P = -ρgh

where h specifically means the difference in vertical position between two points, and P is the pressure difference between those two points. So a better (more explicit) way of expressing this would be:

ΔP = -ρgΔy

which becomes:

P₂ - P₁ = -ρg(y₂ - y₁)

where y₂ is the vertical position of point 2 and y₁ is the vertical position of point 1. If y₂ > y₁, then P₂ < P₁. In other words, the negative sign expresses the fact that pressure increases with depth.

The second problem with your solution is that you are not answering the question that is being asked. It looks like you assumed that the object was partially submerged, and then tried to calculate the weight (not pressure) of the displaced water, and equated it to the weight of the submerged portion of the body. Even then, your calculation was problematic, because the object is a sphere, so its volume is not just simply area x height.

What this question is actually asking is the following: the ball is initially completely submerged (at the bottom of the pool). It begins to rise because there is a net upward buoyant force on it. Eventually it breaks the surface and continues flying upward into the air. How high up into the air does it go? The answer depends upon the speed it has when it reaches the surface. The question asks you to ignore the fact that once the body begins to rise above the surface, the buoyant force goes down. In other words, just pretend that the buoyant force is constant throughout the motion and equal to the value it has when the ball is completely submerged.

So, you can do what you were doing before. Calculate the upward buoyant force, which is just the weight of the displaced volume of water (only you need to use the proper volume for a sphere). Then calculate the downward gravitational force (simple). The difference between these is the NET upward force. Then, ignoring drag (which is highly unrealistic, but whatever), you just have a constant force, which means it is just motion with constant acceleration -- a basic kinematic problem. Since it starts at rest, you can figure out its velocity at the surface, which can be considered the "launch speed." Knowing this launch speed, you can figure out how high into the air it goes.

EDIT: So I guess we didn't really need any of those above formulae for pressure at the beginning of my post, since force is what we were after all along. It's just that you were mistakenly calling the force "pressure" (P). Oh well, I thought it would be best to make the difference between the two clear.