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 slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t (\sqrt{2} + 1) s$ . The value of t is……..(use g = 10 m/s² )

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answer is 2.

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Detailed Solution

From $Energy Conservation = \frac{1}{2} {mv}_{0}^{2} = mgh$

$v_{0} = 10 \sqrt{2}$

For $A \to B$

at B, v = 0

$\begin{array}{l} a = - gsin 45^{\circ} = \frac{- 10}{\sqrt{2}} \\ v = u + {at}_{1} \Rightarrow 0 = 10 \sqrt{2} - \frac{10}{\sqrt{2}} t_{1} \Rightarrow t_{1} = 2 \sec \end{array}$

For $B \to C$

$\begin{array}{l} s = {ut}_{2} + \frac{1}{2} {at}_{2}^{2} \\ \frac{10}{\sin 30^{\circ}} = \frac{1}{2} (10 \sin 30^{\circ}) t_{2}^{2} \\ t_{2} = 2 \sqrt{2} \end{array}$