If a ball fallen freely from 'h' height reaches in time 't' at ground, then what will be the time when it reaches at height `h/2` ?

Explanation

For freefall,` t = sqrt ((2h)/(g)) " "…(1)`
Time taken to reach at `(h)/(2),`
`t . = sqrt ((2h)/(2g)) [ because h . =(h)/(2)]`
` therefore t. = sqrt ((h)/(g)) " "...(2)`
By taking ratio of equation (2) and (1).
`(t.)/(t) = sqrt((h)/(g)xx (g)/(2h)) = (1)/(sqrt2)`
`therefore t . = (t)/(sqrt2) s`