# Basketball pulled underwater then released, what is theoritical velocity and height.

So if you have a typical basketball "mens".

so, my friends and I were in a hot tub and one brought in a basketball.

They were pulling the basketball to the bottom of the hot tub and we noticed that it did not shoot out as high as when we pulled it from the middle.

so we can assume the water has a density of 1. and the basketball is filled to typical values.
we can assume a basketball of 750 mm in circumference. So, V= 4/3* pi * r^3, C= 2*pi*r
r=[(2*pi)/750mm],

V= (4/3)*pi* [(2*pi)/750mm]^3

i don't know what the typical pressure of a basketball is.
and i don't know how many gallons or kilo liters are in our 6 person spa. assume an average amount which i'm having trouble Google'ing

Basically i'm looking for a theoretical explanation of the formulas that would be involved in finding how fast and how high a basketball would shoot out of the water.

The variables i'm thinking could contribute are:
Ball: weight (600 gram), density (89.6 kg/m^3 ), pressure(8 pounds or .51 to .57 atm),
depth.
Water (assume density 1 gram/mL )
:more variables? :
Clearly as the ball is rising up, the water pressure above the ball is decreasing and the pressure below the ball is increasing, much like it hurts on your ears the deeper you swim.

we can assume no gravity i think.

my observation was that the max height and velocity was concave down.
because holding in just underwater will not shoot up very high.
and holding it at the bottom did not "seem" to shoot out as high as just holding it in the middle.

It sounds like a fun problem. Seemingly pointless, but it is about solving it if we can right?

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bump^^^^

You need to use integration for this stuff.

1) Calculate the density of the basketball as a whole.
2) Knowing the density of water, You have to calculate the pressure at the extreme ends of it...since it's a basket ball, i.e it's round upwards pressure will be applied to 50% of the basket ball's area; so actually you have to find an expression for the pressure at every point on the basket ball, then integrate it (with some limit...that I've not thought of) to get the total pressure (or the force on act).
3) Now, using the same procedures you need to calculate the pressure on the rest 50% of the basket ball's area.
3) There will be a difference between the 2 forces computed...this force will the by which the force will be jacked out...but this is not the exact force...since it will reduce as the ball approaches towards the surface of the tub...so AGAIN you have to use integration to calculate something........(I've not thought about it..again); but point is, it will be used to calculate the velocity at which the ball will jack out.
4) Now using laws of motion, you can easily calculate the height which will be very much different from the actual height you'll get since you not work for more complex stuff like the resistance water will pose as the balls move in and the air resistance.

Good luck!

BTW putting a basket ball in a hot tub is not a good idea...it might be hazardous ;)

Try applying this analysis as the first case approximation. I think you can find the bouyancy as a constant proportional to the amount of water displaced by air inside the ball, and ignore pressure differences at shallow depths. This requires solution of the differential equation on a computer or graphing calculator:

http://hyperphysics.phy-astr.gsu.edu/HBASE/lindrg.html#c2

Once the ball hits the air you solve the standard kinematic equation for a vertical projectile with known initial velocity. This equation should predict a maximum height from some starting depth, but not the fall off in maximum height from a deeper depth. I am not sure what would account for that observation ...

Cool i'll look into this, anyone else have any ideas? or simplification. lol

Keep in mind that when doing a fluid dynamics problem not all classical physics apply. I can see two reason why you observed what you did. The ball rising in the water will reach a terminal velocity very quickly, probably within about two feet or so. This would mean that past a certain depth the ball will not be going any faster at the surface no matter how deep you go. Secondly due to the 'no slip' boundary conditions at the surface of the basketball the water is not moving around it, it is sticking to it. As time goes on you start to build up a sheering zone around the ball so that as the ball is moving through the water more of the water is building up on top of the surface of the ball, once it breaks through the air-water interface all of the accumulated water is causing the ball to be heavier, and then you need to consider surface tension which will probably be a significant factor since the area of the water that is disturbed will be fairly large.

So pulling the basketball more than halfway down will cause more water to accumulate above the basketball and slow down the terminal velocity ?

Does anyone know a formula to show the amount of energy the ball is expending getting to the surface? ie how much weight could it pull up and at what speed (I assume this speed would be its eventual terminal velocity, given you put it in something deeper such as the ocean)? Does the amount of energy increase in direct relation to the volume of the basketball, or other larger vessel?

Hi.

When the ball of mass m ,volume V is in water density ρ, depth of the center d, potential energy is -mgd .

When the ball of mass m is still and at height h above the water surface, potential energy is mgh- ρVgd

By conservation of energy, h=(-1 + ρV/m)d

You will find that this is too much simplified, i.e.
- density of air above water is neglected.
- the water surface height of the pool does not change by in/out of the ball.

You can easily improve the answer by considering these details. Regards.

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