Did I calc this right?

I decided to take a shot at DCEU Superman's famed train punch

distance to the train, 300m, here's 100 for reference

time in the air, slightly below 2 seconds - 1.8 s

time Clark's fist made contact with Nam Ek - completely guessing here .01 s

velocity^2 = 29,584 m/s

Nam Ek at over 8 ft I estimate weighs at around 300 kg. the 8'11 Robert Wadlow, without Kryptonian armor and his lanky frame, only weighed in at around 200 kg

△KE = 4,437,600 j - 0

△KE/△T = 4,437,600/.01 = 443,760,000 j or 0.11 tons of tnt

small building levelaccording to these twosites

300m?

The whole town is 19.43km^2 just to contextualize the above.

If Nam's velocity is 300/1.8 = 166.67 m/s how did you get your v^2?

Also why use rate of change of ke? Wouldn't you use the final energy and convert that to tons tnt?

I say below street level striking

@kryptonianking88 said:

I decided to take a shot at DCEU Superman's famed train punch

distance to the train, 300m, here's 100 for reference

time in the air, slightly below 2 seconds - 1.8 s

time Clark's fist made contact with Nam Ek - completely guessing here .01 s

velocity^2 = 29,584 m/s

Nam Ek at over 8 ft I estimate weighs at around 300 kg. the 8'11 Robert Wadlow, without Kryptonian armor and his lanky frame, only weighed in at around 200 kg

△KE = 4,437,600 j - 0

△KE/△T = 4,437,600/.01 = 443,760,000 j or 0.11 tons of tnt

small building levelaccording to these twosites

Not sure how good of an assumption 300m and 300 kg is but assuming those are good:

If he went 300m in 1,8s thats ~167 m/s.

Superman accelerates him to that velocity in ~0.25m.

The contact time is distance divided by average speed so: 0.25/(167/2) = 0.003s.

Acceleration is 167/0.003 = 55.667 m/s².

The force is 55.667 x 300 = 16.700.100 Newtons or 1700 tons of force.

And the energy is = 167^2 x 300 / 2 = 4.183.350 J or 1 kg TNT equivalent.

KE/T doesnt give joules it gives watts/power which is rather useless.

seems about right

@nwname: ah thanks