Posted on Jul 20, 2023
I am hoping this will be a good exercise in math and navigation. How far was the Enola Gay from the detonation of the Little Boy?
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At 08:15 (JST) on 6 August 1945, the Little Boy was released. Col Paul Tibbets said that he pushed the aircraft to its maximum speed.
Use the following to calculate:
Fall time: 44.4 seconds
Maximum speed for a B-29 Superfortress: 357 mph
Release Altitude: 10,000 meters (32,808 feet)
Detonation Altitude: 1968 feet
Use the following to calculate:
Fall time: 44.4 seconds
Maximum speed for a B-29 Superfortress: 357 mph
Release Altitude: 10,000 meters (32,808 feet)
Detonation Altitude: 1968 feet
Posted in these groups: 
Military History
WEPS: Weapons
Military History WEPS: Weapons
Posted >1 y ago
Responses: 1
Lt Col Timothy Cassidy-Curtis
Yes. I got the same answer (I used a spreadsheet). It's about 11 & 3/4 kilometers.
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Lt Col Timothy Cassidy-Curtis
The vertical drop was 32,808 feet, minus 1968 feet, which is 30,840 feet. The horizontal distance was the B-29's maximum speed of 257 mph, which is 523.6 feet per second. The drop time was 44.4 seconds, which allows a flying distance of 523.6 times 44.4, or 23,248 feet. Take the square of each and add. The square root of that sum is 38,016 feet, which is around 7.2 miles.
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LTC Kevin B.
Lt Col Timothy Cassidy-Curtis - I also used a spreadsheet. We had the exact same thing on every step until the final output of the Pythagorean theorem. I had the square root as 38,620.82 feet (which become 7.31 miles). Exact same process to solve though.
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