Let $g$ be the acceleration due to gravity. The problem has up-down symmetry. The time for the object to travel down past the window is $t = T/2$.

Let $v_{top}$ be the object's speed as it passes the top of the window during its descent. Using the kinematic equation for displacement:

$$h = v_{top}t + \frac{1}{2}gt^2$$

Substituting $t = T/2$:

$$h = v_{top}\left(\frac{T}{2}\right) + \frac{1}{2}g\left(\frac{T}{2}\right)^2 = \frac{v_{top}T}{2} + \frac{gT^2}{8}$$

Solving for the speed at the top of the window:

$$v_{top} = \frac{2}{T}\left(h - \frac{gT^2}{8}\right) = \frac{2h}{T} - \frac{gT}{4}$$

This speed is the initial speed for the fall past the window, and also the final speed for the rise from the peak to the top of the window.

To find the maximum height $H$ above the window, we consider the motion from the peak (where velocity is zero) down to the top of the window. Using the kinematic equation $v_f^2 = v_i^2 + 2a\Delta y$:

$$v_{top}^2 = 0^2 + 2gH$$

Solving for $H$:

$$H = \frac{v_{top}^2}{2g}$$

Substituting the expression for $v_{top}$:

$$H = \frac{1}{2g}\left(\frac{2h}{T} - \frac{gT}{4}\right)^2$$