Solution:
The sides $A B$ and $A C$ of the figure are both inclined to the horizontal at $\theta_{1}$ and $\theta_{2}$, respectively. According to the law of mechanical energy conservation,
Potential energy at the top = Kinetic Energy at the bottom
$\therefore \mathrm{mgh}=(1 / 2) \mathrm{mv}_{1}^{2}-(1)$
and $\mathrm{mgh}=(1 / 2) \mathrm{mv}_{2}^{2}-(2)$
Due to the fact that both sides are the same height, both stones will reach the bottom at the same time.
From (1) and (2), we get $\mathrm{v}_{1}=\mathrm{v}_{2}$
So, both the stones will reach the bottom with the same velocity.
For stone 1:
Net force acting on the stone can be calculated as,
$\begin{array}{l}
\mathrm{F}=\operatorname{ma}_{1}=\mathrm{mg} \sin \theta_{1} \\
\mathrm{a}_{1}=\mathrm{g} \sin \theta_{1}
\end{array}$
For stone 2:
$\begin{array}{l}
\mathrm{a}_{2}=\mathrm{g} \sin \theta_{2} \\
\text { As } \theta_{2}>\theta_{1}
\end{array}$
So, $a_{2}>a_{1}$
From $v=u+a t=0+a t$
$\Rightarrow \mathrm{t}=\mathrm{v} / \mathrm{a}$
For stone $1: t_{1}=v / a_{1}$
For stone $2: t_{2}=v / a_{2}$
As $t \propto 1 / \mathrm{a}$, and $\mathrm{a}_{2}>\mathrm{a}_{1}$
So, $t_{2}<t_{1}$
Hence, stone 1 will reach slower than stone 2.
By applying the law of conservation of energy we get
$\mathrm{ngh}=(1 / 2) \mathrm{mv}^{2}$
When the height, $\mathrm{h}=10 \mathrm{~m}$, the speed of the stones is given as
$v=\sqrt{2 g h}=\sqrt{2} \times 9.8 \times 10=14 \mathrm{~m} / \mathrm{s}$