Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block solids down on inclined plane BC. Time it takes to reach to the point C from point A is $t(\sqrt{2}+1) \mathrm{s}$. The value of $t$ is $\ldots \ldots \ldots$. (use $g=10 \mathrm{~m} / \mathrm{s}^2$ )