Answer:
31.32 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
Let us assume the height of the Disque hall is 50 m
[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 50}\\\Rightarrow u=31.32\ m/s[/tex]
In order to make the jump Superman's initial velocity must be greater than or equal to 31.32 m/s
