Maximum Acceleration for Moon Launch

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The discussion revolves around calculating the maximum acceleration experienced by space travelers during a hypothetical Moon launch proposed by Jules Verne. Participants initially struggle with the correct application of the kinematic equation, ultimately needing to express acceleration as a multiple of gravitational acceleration (g = 9.81 m/s²). The correct approach involves using the final velocity of 10.78 km/s and the distance of 220 m to derive acceleration accurately. There is confusion regarding the correct formulation of the equations, leading to repeated corrections and clarifications. The conversation concludes with a participant indicating they have resolved the problem before the deadline.
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In 1865, Jules Verne proposed sending men to the Moon by firing a space capsule from a 220 m long cannon with final velocity of 10.78 km/s. What would have been the unrealistically large acceleration experienced by the space travelers during launch?

The answer needs to be in

Number times g.
g = 9.81

I tried v^2 = 1/2(a)(220)
So v^2 = 110
Square root of 110 = 10.48..
Not correct..

Don't know what to do.
 
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Kildars said:
I tried v^2 = 1/2(a)(220)
So v^2 = 110
Square root of 110 = 10.48..
Not correct..

Don't know what to do.
Don't throw away the (a) going from the first to the second line, and use the given information to calculate v^2. Then solve for (a) and express as a fraction of g. Also, check your first equation. It is not correct.
 
Last edited:
OlderDan said:
Don't throw away the (a) going from the first to the second line, and use the given information to calculate v^2. Then solve for (a) and express as a fraction of g. Also, check your first equation. It is not correct.


Right so 10,780^2 is the final velocity since the first one is in k/m?

10,780^2 = 1/2(a)(220)
10,780^2 = 110a
A = 1056440
Divided by 9.81..

107690.1121

Still incorrect.
 
Kildars said:
Still incorrect.
Perhaps because your equation is still incorrect, and you have not expressed a as a multiple of g.
 
OlderDan said:
Perhaps because your equation is still incorrect, and you have not expressed a as a multiple of g.

Can I not just divide the acceleration by g at the end and get the same answer?
 
bump..due in an hour 20 minutes ;)
 
Nevermind, got this one.
 
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