A block is released from top of a smooth inclined plane It reaches the bottom of the plane in `sqrt2s` The time taken (in second) by the body to cover Ist half of inclined plane is .

Answer

`1`

Explanation

Let `2l` be the length of inclined plane and `0` be the angle of inclination as show in
From `s = ut + (1)/(2)at^(2)`
`2l = 0+ (1)/(2) (g sin theta)t^(2)`
Dividing (i) by (ii) we get
`2 = (2)/(t^(2)), t^(2) = (2)/(2) =1, t =1s`
.