#### Answer

The ball will reach a height of 2h

#### Work Step by Step

The potential energy at the maximum height is equal to the kinetic energy at the initial position.
$PE_1 = KE_0$
$mgh = KE_0$
$h = \frac{KE_0}{mg}$
Let's suppose that the kinetic energy is $2KE_0$ at the initial position and let $h_2$ be the height the ball reaches.
$PE_2 = 2KE_0$
$mg~h_2 = 2KE_0$
$h_2 = \frac{2KE_0}{mg}$
$h_2 = 2~\frac{KE_0}{mg}$
$h_2 = 2~h$
The ball will thus reach a height of 2h.