The Nuclear Bomb That Was Dropped off the Coast of Georgia

On February 5, 1958, the United States Air Force dropped a 7,600-pound nuclear bomb into the water off the coast of Georgia. The crew of the B-47 bomber didn’t drop the bomb by accident. They dropped it in order to survive.

The B-47 bomber was flying a top-secret simulated combat mission from Homestead Air Force Base in Florida. The airplane was carrying a single Mark 15 thermonuclear bomb to make the training as realistic as possible. The bomb contained highly enriched uranium and 400 pounds of high explosives. The plutonium core that would have triggered a nuclear reaction had been replaced with a dummy lead core.

F-86 fighter aircraft were also involved in the exercise and were tasked with intercepting the bomber. Something went terribly wrong, however, when one of the fighter jets didn’t see the bomber on radar. The fighter jet collided with the B-47 at 38,000 feet and knocked out its engine and put a massive hole in the wing. The F-86 had its left wing sheared off.

Colonel Howard Richardson was at the bomber’s controls and managed to get the aircraft leveled out at 20,000 feet, but he had to make a decision. He decided the best course of action for him and his fellow crew member was to jettison the heavy load they were carrying to make the aircraft lighter for an emergency landing as well…