$A$ uniform sphere of mass $m$ and radius $R$ is placed on a rough horizontal surface. The sphere is struck horizontally at a height $h$ from the floor. Match the following:

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

(A-(III), B-(IV), C-(II), D-(I)) When a sphere is struck at height $h$,the impulse $J$ provides linear momentum $mv = J$ and angular momentum about the center $L = J(h - R) = I\omega = \frac{2}{5}mR^2\omega$.
Thus,$v = \frac{J}{m}$ and $\omega = \frac{J(h-R)}{\frac{2}{5}mR^2} = \frac{5J(h-R)}{2mR^2}$.
The velocity of the bottom point is $v_{bottom} = v - \omega R = \frac{J}{m} - \frac{5J(h-R)}{2mR} = \frac{J}{2mR} [2R - 5h + 5R] = \frac{J}{2mR} [7R - 5h]$.
For pure rolling,$v_{bottom} = 0$,which implies $h = \frac{7R}{5}$. This corresponds to $(d)-(i)$.
If $h < \frac{7R}{5}$,$v_{bottom} > 0$,so the sphere slips forward,friction acts backward,causing clockwise spin (relative to center),which corresponds to $(ii)$ or $(iv)$. Specifically,if $h=R$,$\omega=0$,so it has only translational motion $(iv)$. If $h < R$,$\omega$ is negative (anti-clockwise),so $(iii)$. If $R < h < \frac{7R}{5}$,$\omega$ is positive (clockwise),so $(ii)$.
Thus: $(a)-(iii)$,$(b)-(iv)$,$(c)-(ii)$,$(d)-(i)$.