Pool Ball Terminal Velocity

I just terminated a duplicate post with a fair amount of velocity.
 
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Cornerman said:
About 1.5 second drop, goes about 35 ft.
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What if I took a ball and hung it from a 35ft string, then dropped it like a pendulum so it would "swing down" instead of fall.

How much difference would that make in the velocity, once it has dropped the 35ft, as opposed to a free fall.
 
Mr. Bond said:
What if I took a ball and hung it from a 35ft string, then dropped it like a pendulum so it would "swing down" instead of fall.

How much difference would that make in the velocity, once it has dropped the 35ft, as opposed to a free fall.
Click to expand...

No difference

In either case all of what was previously potential energy (the mass lifted up 35 feet) is converted to kinetic energy--energy of motion.
 
BC21 said:
Using the terminal velocity equation below, and assuming a normal air density of 1.27kg/m3, and a drag coefficient of 0.5 for a sphere.....



V = √(2mg/pAc)



m = mass of CB = 170.1g

g = gravity = 9.8m/s2

c = drag coefficient = 0.5

A = area of falling object = 25.66cm2 (convert to meters = 0.002566m2)

p = air density = 1.27kg/m3 (covert to grams3 = 1,270g/m3)





Plug it all in and get 45.24m/s, or 101.2mph. Nobody can break the balls that fast!
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Yeah not even close lol. What's the record maybe 35 mph


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The post asked with no air resistance - in a vacuum. an academic question.

How about a tail wind to offset the air resistance?
 
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