Vertical Travel of a projectile
Vertical Travel of a projectile
(OP)
I have a threaded bar which could act as a sealed plug being forced vertically out due to pressure.
F = x-section area x pressure
I think I can calculate the initial accelaration as per,
F = ma - mg
I am trying to find its maximum vertical travel.
F = x-section area x pressure
I think I can calculate the initial accelaration as per,
F = ma - mg
I am trying to find its maximum vertical travel.





RE: Vertical Travel of a projectile
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RE: Vertical Travel of a projectile
RE: Vertical Travel of a projectile
David
RE: Vertical Travel of a projectile
F = ma
v = at
How much t?
RE: Vertical Travel of a projectile
-handleman, CSWP (The new, easy test)
RE: Vertical Travel of a projectile
So you have a mass with unbalanced force due to pressure upwards, fighting gravity. This is a general dynamics question, you can use various equations to get the height of the projectile.
But I think you got your acceleration wrong at the start of the problem.
Regards,
Cockroach
RE: Vertical Travel of a projectile
As noted - this is a simple physics problem.
I made a potato gun and my buddy and I wanted to find the potato speed. So - using basic and "rough" physics we decided to shoot it straight up and time it. Speed up will be about speed coming down or at least fairly close.
So that's what we did. Must have aimed it fairly well because after it hit apogee and started coming down - it was quite obvious we might get hit!!
We ran like scared school kids. Sure enough - it hit within 1' from the launch site!!
If memory serves - it was over 80 mph.
RE: Vertical Travel of a projectile
Weight of the rod, W = mg
Force of the gas, F = PA
Xsec area of the rod, A
Pressure of the gas, P
Distance to travel to exit the bore, d
Height traveled to v = 0, h
Exit velocity, v
When rod exits the bore, (F-W)*d = 0.5*m*v^2, work done = kinetic energy at exit.
When rod reasches h, W*h = 0.5*m*v^2, potential energy at h = KE at exit.
therefore, W*h = (F-W)*d, or h = (F-W)*d/W
Assume pressure is constant during rod travel through distance d, then F = P*A
Then, h = (P*A-W)*d/W
Ted
RE: Vertical Travel of a projectile
-handleman, CSWP (The new, easy test)
RE: Vertical Travel of a projectile
This is not a potato gun, and some of you won't need a helmet while going to work. There are enough resistances to the force created by the pressurisation, besides there is a pressure equalisation design which takes the force created to a considerably lower value.
I have got my answer, thank you all.