Two inclined planes are placed as shown in figure. A block in 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 10m. 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 $t (\sqrt[]{2} + 1) s$ . The value of t is__________.

(use g = 10 m/s²)

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Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

From conservation of Energy

$\frac{1}{2} m v_{0}^{2} = m g h$

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

For A to B

$a = - g s i n 45 ° = \frac{- 10}{\sqrt[]{2}}$

T = t₁ + t₂

$\begin{array}{l} = 2 \sqrt[]{2} + 2 \\ = 2 (\sqrt[]{2} + 1) \end{array}$

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