Two inclined planes are placed as shown in th | vedclass

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(C) $1$. Motion from $A$ to $B$: The block is projected with velocity $u$ such that it just reaches $B$ $(v=0)$. By conservation of energy, $\frac{1}{

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$2$

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(C) $1$. Motion from $A$ to $B$: The block is projected with velocity $u$ such that it just reaches $B$ $(v=0)$. By conservation of energy, $\frac{1}{2}mu^2 = mgh$. Thus, $u = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 10} = 10\sqrt{2} \ m/s$.$2$. The acceleration along the incline $AB$ is $a_1 = -g \sin 45^{\circ} = -\frac{10}{\sqrt{2}} \ m/s^2$.$3$. Using $v = u + a_1t_1$, we get $0 = 10\sqrt{2} - \frac{10}{\sqrt{2}}t_1$, which gives $t_1 = 2 \ s$.$4$. Motion from $B$ to $C$: The block starts from rest $(u=0)$ and slides down incline $BC$. The length of $BC$ is $L = \frac{h}{\sin 30^{\circ}} = \frac{10}{0.5} = 20 \ m$.$5$. The acceleration along $BC$ is $a_2 = g \sin 30^{\circ} = 10 	imes 0.5 = 5 \ m/s^2$.$6$. Using $s = ut_2 + \frac{1}{2}a_2t_2^2$, we get $20 = 0 + \frac{1}{2}(5)t_2^2$, so $t_2^2 = 8$, which means $t_2 = 2\sqrt{2} \ s$.$7$. Total time $T = t_1 + t_2 = 2 + 2\sqrt{2} = 2(\sqrt{2} + 1) \ s$.$8$. Comparing with $t(\sqrt{2}+1)$, we find $t = 2$.

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