A particle is placed at the point A of a frictionless track ABC as...

Applying conservation of mechanical energy between point $A$ & $B$, we get

$K{E}_{A} + {U}_{A} = K{E}_{B} + {U}_{B}$

$⇒ 0 + mg (1)= \dfrac{1}{2}m{v}^{2} + mg \times 0 . 5$

$⇒ v = \sqrt{g} = \sqrt{10}m{s}^{- 1}$