The Physics of Superheroes: Spectacular Second Edition by James Kakalios

change in speed over time and that velocity is the change in distance over time. When the dust settles, we find that the relationship between Superman’s initial velocity v and the final height h of his leap is v × v = v 2 = 2gh. That is, the height Superman is able to jump depends on the square of his liftoff velocity, so if his starting speed is doubled, he rises a distance four times higher.
    Why does the height that Superman can leap depend on the square of his starting speed? Because the height of his jump is given by his speed multiplied by his time rising in the air, and the time he spends rising also depends on his initial velocity. When you slam on your auto’s brakes, the faster you were driving, the longer it takes to come to a full stop. Similarly, the faster Superman is going at the beginning of his jump, the longer it takes gravity to slow him down to a speed of zero (which corresponds to the top of his jump). Using the fact that the (experimentally measured) acceleration due to gravity g is 32 feet per second per second (that is, an object dropped with zero initial velocity has a speed of 32 feet/sec after the first second, 64 feet/sec after the next second, and so on) the expression v 2 = 2gh tells us that Superman’s initial velocity must be 205 feet/sec in order to leap a height of 660 feet. That’s equivalent to 140 miles per hour! Right away, we can see why we puny Earthlings are unable to jump over skyscrap ers, and why I’m lucky to be able to leap a trash can in a single bound.
    In the above argument, we have used Superman’s average speed, which is simply the sum of his starting speed (v) and his final speed (zero) divided by two. In this case his average speed is v/2, which is where the factor of two in front of the gh in v 2 = 2gh came from. In reality, both Superman’s velocity and position are constantly decreasing and increasing, respectively, as he rises. To deal with continuously changing quantities, one should employ calculus (don’t worry, we won’t), whereas so far we have only made use of algebra. In order to apply the laws of motion that he described, Isaac Newton had to first invent calculus before he could carry out his calculations, which certainly puts our difficulties with mathematics into some perspective. Fortunately for us, in this situation, the rigorous, formally correct expression found using calculus turns out to be exactly the same as the one obtained using relatively simpler arguments, that is, v 2 = 2gh.
    How does Superman achieve this initial velocity of more than 200 feet/sec? As illustrated in fig. 5, he does it through a process that physicists term “jumping.” Superman crouches down and applies a large force to the ground, and the ground pushes back (since forces come in pairs, according to Newton’s third law). As one would expect, it takes a large force in order to jump up with a starting speed of 140 mph. To find exactly how large a force is needed, we make use of Newton’s second law of motion, F = ma —that is, Force is equal to mass multiplied by acceleration. If Superman weighs 220 pounds on Earth, he would have a mass of 100 kilograms. So to find the force, we have to figure out his acceleration when he goes from standing still to jumping with a speed of 140 mph. Recall that acceleration describes the change in velocity divided by the time during which the speed changes. If the time Superman spends pushing on the ground using his leg muscles is ¼ second, 12 then his acceleration will be the change in speed of 200 feet/sec divided by the time of ¼ second, or 800 feet/sec 2 (approximately 250 meters/sec 2 in the metric system, because a meter is roughly 39 inches). This acceleration would correspond to an automobile going from 0 to 60 mph in a tenth of a second. Superman’s acceleration results from the force applied by his leg muscles to get him airborne. The point of F = ma is that for any change in motion, there must be an applied force and the