- #1

freydawg56

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So if you have a typical basketball "mens".

http://en.wikipedia.org/wiki/Basketball_(ball )

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?

http://en.wikipedia.org/wiki/Basketball_(ball )

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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