In action comics #1 (1938), superman could "leap tall buildings in a single bound." initially, superman was said to have come from a planet with a larger gravity than earth's gravity and that was the sole source of his "powers". Based on this premise and the fact that in 1938 a tall building was about 40 stories (122 meters), let's try to answer the following questions about superman's leaping of tall buildings: a. How long would he be in the air? b. How fast would Superman be by launching himself off the ground? c. What acceleration would he have to produce to achieve this launch speed?

[tex]v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 122}\\\Rightarrow u=48.92\ m/s[/tex]

The speed at which Superman launches himself is 48.92 m/s

[tex]v=u+at\\\Rightarrow 0=48.92-9.81t\\\Rightarrow t=\frac{-48.92}{-9.81}=4.98\ s[/tex]

An object which is thrown at an initial velocity when it reaches the same point from where it was thrown it achieves the same initial velocity

[tex]v=u+at\\\Rightarrow 48.92=0+9.81t\\\Rightarrow t=\frac{48.92}{9.81}=4.98\ s[/tex]

Time to come down is 4.98 seconds

Total time Superman will be in the air is 4.98+4.98 seconds = 9.96 seconds

[tex]v=u+at\\\Rightarrow 48.92=0+a\\\Rightarrow a=\frac{48.92}{t}\ m/s^2[/tex]

Here, the time taken to accelerate to his initial speed is not given. It can be assumed that it is almost instantaneous. The time may be infinitesimally small. So, the acceleration might be almost 48.92 m/s²