Two inclined planes are placed as shown in figure. A block is...

From energy conservation at point $A$ and $B$,

$\dfrac{1}{2}m{v}_{0}^{2} = mgh$

$⇒ {v}_{0} = \sqrt{gh} = \sqrt{10 \times 10}$

$⇒ {v}_{0} = 10\sqrt{2}m{s}^{- 1}$

For $A \to B$

At $B, v = 0$

Acceleration of a particle moving on a smooth incline is $gsin\theta$. Therefore, along $AB$, $a = - gsin45 ° = \dfrac{- 10}{\sqrt{2}}m{s}^{- 2}$.

Using equation of motion,

$v = u + a{t}_{1}$

$0 = 10\sqrt{2} - \dfrac{10}{\sqrt{2}}{t}_{1} ⇒ {t}_{1} = 2s$

For $B \to C$

Using second equation of motion,

$s = u{t}_{2} + \dfrac{1}{2}a{t}_{2}^{2}$

$⇒ \dfrac{10}{\sin 30 °} = \dfrac{1}{2}\left(10\sin 30 °\right){t}_{2}^{2}$

${t}_{2} = 2\sqrt{2}s$

So total time $T = {t}_{1} + {t}_{2}$ $= 2\sqrt{2} + 2$

$= 2\left(\sqrt{2} + 1\right)s$