Where to kick a ball to achieve rolling during the whole motion? That's all the information I've got, so I don't really know what to do. But this is my idea: The equations of the motion are
$$ma=F_0\delta(t)-F_{friction}$$ and $$\frac{2mR^2}{5} \frac{d \alpha}{d t}=F_{friction}R+\delta(t)F_0 r,$$ where $R$ is the radius of the ball, $m$ is the mass of the ball and $r$ is the distance from the center of the ball to the hitting point. But I think I've done it wrong.
 A: Suppose you have a motionless ball of radius $R$ resting on a horizontal plane with finite friction coefficient $\mu$. You kick the ball at a height $h$ above the ground along the horizontal direction and observe its motion. You want to find $h$ where it starts with pure rolling, instead of sliding.
This is the axis of percussion problem, and it is fairly simple to solve.

What is important is the mass of the ball $m$, and the mass moment of inertia about the center of mass $I$. 
The impulse $J$ from the kick might cause and reaction impulse $G$ from the ground, and the friction is low this reaction is limited in magnitude due to traction before the ball slips. So the goal here is to find $h$ such that $G=0$.
The equations of motion (horizontal and rotational) are:
$$ \begin{aligned}
  J - G & = m v \\
  J (h-R) + G R & = I \omega \\
\end{aligned} $$
and the pure rolling condition is $v = \omega R$
The solution to the above two equations is
$$ \begin{aligned}
  \omega &= \frac{h}{I+m R^2} J \\
  G & = J - \frac{h\,R\,m}{I+m R^2} J
\end{aligned} $$
To achieve the goal of $G=0$, solve the 2nd equation for $h$
$$ \boxed{ h = R + \frac{I}{m\,R} } $$
The resulting motion is thus
$$ \begin{aligned}
\omega &= \frac{J}{m\,R} & v & = \frac{J}{m}
\end{aligned} $$
NOTE: The axis of precussion defined by $h$ is always above the center of mass, and that makes sense since if it was near the center of mass the ball would try to translate (and slip).
References:


*

*The same math for a related problem although inverted.

*An even more detailed analysis, of how to efficiently slide a bowling pin (the opposite problem).

