After pondering this some more and trying a different method to work out the energy for catapulting , I have the following thoughts.
The ball shows 7.1 lbs when I weigh it with a scale. I had been treating 7.1 lbs as the mass of the ball for purpose of catapulting it… i.e. accelerating it to a particular velocity to be able to go to a height to acheive the Hurdle. That is because the scale does not show (most of) the mass of air in the ball. Yes it is true that the external atmosphere largely cancels the “weight” effect of the mass of air in the ball because the air in the ball is likely to be only 1 psi greater than atmospheric pressure. On the surface, it would seem that we still need to accelerate the entire ball mass mass and that would seemingly take more energy than accelerating the ball only.
From the above post, assuming 1 psig inflation, I estimate that the mass of air in the ball that is immune to the “net” effect of gravity (invisible to my scale) is 0.623 kg or 1.37 lbs. Net affect of gravity is considering the effect of buoyancy of the atmosphere acting on the ball. 0.091 lbs weight is from the overpressure does see the “net” effects of gravity. This 0.091 lbs is included in the 7.1 lb weight of the ball when I weighed it. i.e. the uninflated ball should weigh 7.01 lbs
Thus the mass of air in the 1 psig ball is 1.37+ 0.091 lbs = 1.461 lbs mass. This is 17.2% of the mass of the 8.47 lb ball.
Is it possible that the mass of the buoyant (zero weight) air in the ball is not a real issue for the purpose of calcuating hurdling height,velocity, energy? The buoyant air I speak of is the 1.37 lbs at 1 atm and not the 0.091 lbs mass due to the inflation pressure of 1 psi over atm. .
I am thinking that once the buoyant (zero weight ) 1.37 lbs of air is accelerated, it has kinetic energy… yet it is immune to gravity. As such, this kinetic energy is eventually depleted to propel upwards the components that have the net weight. Is this why punching a day old helium balloon upwards results in a slow motion parabolic trajectory (including ascent, and ignoring drag coefficient)?
Bottom line is that through the effect of bouyancy of our atmosphere, an 8.47 lbs mass ball has a weight of 7.1 lbs. It is like gravity has been reduced by 1- (7.1/8.47) or ~16.1% to 83.9% of g.
It appears that in the end, we do not need extra velocity for propelling the total mass of 8.47 lbs (vs velocity needed for 7.1 lbs) to get the height desired. Relations are: PE=mgh , KE=1/2 mv^2 —> h= (v^2)/2g, v=sqrt(2gh) where v = vlaunch.
This is as long as the mass is taken as the 8.47 lbs and g is corrected by a factor of (ball weight)/(ball mass) = 0.83825.
In fact, launch velocity of the ball is reduced compared to a scenario where there was no bouyant atmosphere and the mass taken as 8.47 lbs.
Sure some will observe that we could have ignored the bouyant mass for calculating the energy required to propel the ball to the desired height. This is true. The launch velocity and time of travel in teh kinematics calcs would however be incorrect if you do not use the above method of correcting g. Yes we have ignored air drag.
Hmm, the ball gets a bit of the slow mo effect that you would have on the moon.
Appologies for not doing this in metric as well.