Solution:

Let the initial velocity of the ball be \( u \), and let \( v \) be the velocity of the ball at a distance \( h \) above the tower. Using the equation of motion:

\[
v^2 = u^2 - 2gh
\]

At a distance \( h \) below the tower, the velocity is doubled, so:

\[
(2v)^2 = u^2 + 2gh
\]

Simplifying these:

\[
4v^2 = u^2 + 2gh
\]

Substitute \( v^2 = u^2 - 2gh \) from the first equation:

\[
4(u^2 - 2gh) = u^2 + 2gh
\]

Expanding and solving:

\[
4u^2 - 8gh = u^2 + 2gh
\]

\[
3u^2 = 10gh
\]

\[
u^2 = \frac{10gh}{3}
\]

The maximum height \( H \) from the top of the tower is given by:

\[
H = \frac{u^2}{2g} = \frac{10gh}{6g} = \frac{5h}{3}
\]

Thus, the maximum height attained by the ball from the top of the tower is \( \frac{5h}{3} \).