$A$ body falls freely from a height $h$ in a gravitational field. Match the following:

Column-$I$	Column-$II$
$(1)$ Potential Energy $=$ Kinetic Energy	$(a)$ At height $\frac{2h}{3}$
$(2)$ Potential Energy $= 2 \times$ Kinetic Energy	$(b)$ Constant at every point
	$(c)$ At height $\frac{h}{2}$

(A) Let the total energy be $E = mgh$. At any height $y$ from the ground,Potential Energy $U = mgy$ and Kinetic Energy $K = mg(h-y)$.
For $(1)$,$U = K \implies mgy = mg(h-y) \implies y = h-y \implies 2y = h \implies y = \frac{h}{2}$. Thus,$(1) \rightarrow (c)$.
For $(2)$,$U = 2K \implies mgy = 2mg(h-y) \implies y = 2h - 2y \implies 3y = 2h \implies y = \frac{2h}{3}$. Thus,$(2) \rightarrow (a)$.
Therefore,the correct matching is $(1)-(c)$ and $(2)-(a)$.