A particle is released from a height H . At a certain height its kinetic energy is half of its potential…

Given, $P.E.=2K.E.$

now conservation of kinetic energy and potential energy,

$$\begin{gathered} P.E.+K.E.=mgH \\ \Rightarrow P.E.+\left(\frac{1}{2}\right)P.E.=mgH \\ \Rightarrow \frac{3}{2}P.E.=mgH \\ \Rightarrow P.E.=\frac{2mgH}{3}=mgh \\ \Rightarrow h=\frac{2H}{3} \end{gathered}$$

here the kinetic energy,

$$\begin{gathered} 2K.E.+K.E.=mgH=3K.E.=3\left(\frac{1}{2}m{v}^2\right) \\ v=\sqrt{\frac{2gH}{3}} \end{gathered}$$