## Space Cannon to the Moon

1. The problem statement, all variables and given/known data
What are the acceleration of the cannon and the g-forces experienced if there was a guy on it...

Cannon is 215m long

Velocity is 1.66x10^4 ms^-2

2. Relevant equations
V2=u2 + 2ay

3. The attempt at a solution
(1.66x104)^2=0 + 2xax215

a = (1.66x104)[SUP]^2/2x215

Ok... So i think that's the acceleration?? I'm not sure.
I am not completely stuck on how to the g-force...

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 G force is normally acceleration compared to the earths gravity we experience on the surface. So if you experience 4 G's then you're being accelerated at 4*9.81 m/s^2.

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hi midgetwars! welcome to pf!

(try using the X2 icon just above the Reply box )

yes, that's fine

the acceleration (plus 1g) is the g-force

(yes, i know it's called a force … but it isn't! )

see http://en.wikipedia.org/wiki/G-force for some details, including …
 The g-force (with g from gravitational) on something is its acceleration relative to free-fall.
btw, are you sure it's 215 m … that's a very long cannon!

## Space Cannon to the Moon

So the a is

so i got 26 1302.3256 ms0-2...

That means the g-forces are around 26 000 gs??? That's seems to be a lot don't you think?? Also wikipedia says the acceleration isn't that high, but doesn't offer a solution

a = (1.06x104)2 / 2x215

 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor how did you get that? EDIT: oh you've just changed it
 So is that right?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor a = (1.66x104)2/2x215 is right

 Quote by tiny-tim a = (1.66x104)2/2x215 is right

so the G forces would be + 1...?

 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor + g, yes

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