Answer: 3.0 m
Using the equation of motion:
[tex]v^2-u^2=2as[/tex]
where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance covered.
[tex] v=0m/s\\ \Rightarrow -u^2=2(-g)\times 0.5=g\\ \Rightarrow u^2=g---1[/tex]
(g is the acceleration due to gravity on Earth)
we know, acceleration due to gravity on moon is one-sixth the value on earth.
Using the value of initial velocity calculated in equation 1, we can calculate the height attained on the moon when an astronaut would jump on moon with the same initial velocity. At the highest point, the instantaneous velocity v is 0 m/s.
[tex]s=\frac{v^2-u^2}{2a}=\frac{-g}{2(-g/6)}=\frac{6}{2}=3.0\hspace{1mm}m[/tex]
