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Moon Punching Question

We already have the speed because it’s 384,472,282 meters from the Earth to the moon, divided by 54 seconds=7,119,857.074074074 m/s. I think.
 
Assuming the average person is 62 kg...

0.5*62*7,119,857.074074074^2= 1.5714633074125216559159376076818e+15 J or 375.59 kiloton of TNT (Large Town level)

However, we recently had a revision for kinetic energy calcs where carrying a human being or sending other human-sized characters flying into the air might not be usable unless there's additional context.
 
Assuming the average person is 62 kg...

0.5*62*7,119,857.074074074^2= 1.5714633074125216559159376076818e+15 J or 375.59 kiloton of TNT (Large Town level)

However, we recently had a revision for kinetic energy calcs where carrying a human being or sending other human-sized characters flying into the air might not be usable unless there's additional context.
What context do we need?
 
For some reason I can't find it anymore, it must have been deleted from the original forums during the Forum move, maybe you could try out the blog posts where there are archives of said revision thread.
 
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