Two inclined planes are placed as shown in figure. A block is projected from the point \( A \) o...

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

[nadanshaks]

Par main tho solution dekhne aaya tha add hi chal raha hai 2 min se 😂