We can use equations of motion here.
`S = ut + (1/2) at^2`
`s = h`
`u = 0`
`t = ?`
`a = g`
`h = (1/2)*g(t)^2`
`t = sqrt((2h)/g)` since `t>0`
So the ball will be dropped to ground after `sqrt((2h)/g) ` time.
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