After falling from a height of 200 m , water flows horizontally with a certain velocity . Ignoring any energy dissipation , find the velocity of flow.

Explanation

Potential energy changes into kinetic energy during the free fall of water.
Let mass of water =m, height =h , final velocity =v
Here, at the height of 200 m
potential energy `(P.E)_i` = mgh and kinetic energy `(K.E)_i=0`
When water falls and flows horizontally,
potential energy , `(P.E)_f =0` and kinetic energy `(K.E.)_f=1/2 mv^2`
`therefore` From conservation of mechanical energy we have,
`therefore mgh =1/2 mv^2`
`or, v=sqrt(2gh) =sqrt(2xx9.8xx200)`
`=62.61 m*s^(-1)`.