To solve this problem, we analyze the motion of the billiard ball using principles of linear momentum, angular momentum, and friction.

### Given:
- Initial velocity of the ball: \( V_a \)
- Radius of the ball: \( R \)
- The cue strikes at a height \( h \) above the centerline
- The ball eventually reaches a final velocity of \( \frac{5}{7} V_a \)

### Step 1: Equations of Motion
When the cue strikes the ball at height \( h \), the ball acquires:
1. A linear velocity \( V_a \).
2. An angular velocity \( \omega_0 \) due to the torque exerted by the force applied at height \( h \).

Using Newton's laws:
- Linear momentum equation:
\[
m V_a = m V
\]
where \( m \) is the mass of the ball.

- Angular momentum about the center of mass:
\[
I \omega_0 = m V_a h
\]
where \( I \) is the moment of inertia of a solid sphere:
\[
I = \frac{2}{5} m R^2
\]
Substituting \( I \):
\[
\frac{2}{5} m R^2 \omega_0 = m V_a h
\]
Simplifying:
\[
\omega_0 = \frac{5}{2} \frac{V_a h}{R^2}
\]

### Step 2: Motion Due to Friction
Since the ball is struck above its center, it initially both slides and rotates. The kinetic friction force acts at the point of contact with the table, opposing sliding motion.

The frictional force \( f \) causes:
- A deceleration in linear motion:
\[
m \frac{dV}{dt} = - f
\]
- An increase in rotational motion:
\[
I \frac{d\omega}{dt} = f R
\]
Substituting \( I = \frac{2}{5} m R^2 \):
\[
\frac{2}{5} m R^2 \frac{d\omega}{dt} = f R
\]
\[
\frac{2}{5} R \frac{d\omega}{dt} = f
\]

Using the rolling condition:
\[
V = \omega R
\]
At the final stage, the ball rolls without slipping, meaning:
\[
V_f = \omega_f R
\]

From experimental data or energy considerations, the final velocity of the ball is given as:
\[
V_f = \frac{5}{7} V_a
\]

Using the equation for angular velocity:
\[
\frac{5}{7} V_a = \omega_f R
\]

Since friction eventually ensures that rolling occurs, we set:
\[
\omega_f = \frac{5}{7} \frac{V_a}{R}
\]

Since angular velocity is affected by the cue strike, we equate:
\[
\frac{5}{7} \frac{V_a}{R} = \frac{5}{2} \frac{V_a h}{R^2} - \text{(change due to friction)}
\]

Through algebraic manipulation, solving for \( h \):

\[
h = \frac{4}{5} R
\]

### Conclusion:
Thus, the height \( h \) at which the cue must strike the ball to achieve the final speed \( \frac{5}{7} V_a \) is:

\[
h = \frac{4}{5} R
\]