Answer:
3.6 m/s
Explanation:
We are given that
Initial velocity of free falling body,u=0
Distance, h[tex]=0.663 m[/tex]
We have to find the velocity of an untethered stone after falling 0.663 m
[tex]v^2-u^2=2gh[/tex]
Where [tex]g=9.8m/s^2[/tex]
Substitute the values
[tex]v^2-0=2\times 9.8\times 0.663[/tex]
[tex]v=\sqrt{2\times 9.8\times 0.663}[/tex]
[tex]v=3.6 m/s[/tex]
Hence, the velocity of an untethered stone after falling 0.663 m=3.6 m/s