Super heroes are great and they normally make great movies (the Fantastic Four movies do not count). One of my favorite heroes of all time is Superman. I mean who wouldn't want to be able to fly or have super strength. Going back to the original Superman Comics he was not as strong as he is today only able to jump a measly 660 feet in the air, but that is still incredible. I was wondering just how much force it would take to allow Superman to jump like that. After some research I found this book titled "The Physics of Superheroes: Second Edition." It has a plethora of different examples of physics of superheroes and one of the chapters has my personal favorite Superman. The talks about how Superman jumps. The author, a physics professor himself, goes on to imply even superman follows the basic principal that force equals mass multiplied by acceleration. Then the author goes on to talk about the original Superman's jump of 660 feet saying that gravity is the main cause of the jump stopping ( Chapter 1 TPofSH). After some math the author says that to achieve the amazing jump height his starting velocity needs to be squared from his takeoff acceleration which is 200 feet a second which is equal to 140 miles per hour. Then to get the force applied to the ground he assumes Superman is 100 kilograms. The time assumed to jump off the ground is about a fourth of a second so he multiplied the 200 feet a second to account for time which equals 250 meters per second squared after also converting to the metric system. So multiplying Superman's mass by his acceleration gives a whopping 25000 meters a second squared. When Superman jumps he applies a whopping 5600 pounds of force to the nearby ground.
Kaklios, John. "Chapter 1." The Physics of Superheroes: Spectacular Second Edition. N.p.: Penguin, 2009. N. pag. Print.