5. Since it is required that the stone does not hit the specified segment, the sought interval of velocities has the form $v>v_{0}$. Let's call $v_{0}-$ the minimum velocity.

By choosing the throwing point and the angle of inclination of the initial velocity to the horizon, one can ensure that the trajectory passes through any two given points not lying on the same vertical. From the law of conservation of energy, it follows that the case of minimum velocity will correspond to a trajectory passing through both corners of the building.

Let the stone, passing through the upper part of the building, have a velocity $v_{1}$, directed at an angle $\beta$ to the horizon. The geometric locus of points through which all possible trajectories pass is described by the inequality

$$
y \leq \frac{v_{1}}{2 g}-\frac{g x^{2}}{2 v_{1}^{2}}
$$

in a coordinate system with the origin at the top of the building and axes $x y$ directed horizontally in the direction of the slope and vertically (the safety parabola).

The smallest value of $v_{1}$ corresponds to the case

$$
-l \operatorname{tg} \alpha=\frac{v_{1}}{2 g}-\frac{g l^{2}}{2 v_{1}^{2}}
$$

In this case, the trajectory passes through both vertices of the building and, due to the convexity of the parabola, does not intersect the roof. From this,

$$
v_{1}^{2}=g l\left(\sqrt{\operatorname{tg}^{2} \alpha+1}-\operatorname{tg} \alpha\right)=g l \frac{1-\sin \alpha}{\cos \alpha}
$$

By the law of conservation of energy, the minimum velocity of the stone on the ground is

$$
v_{0}=\sqrt{g\left(2 H+l \frac{1-\sin \alpha}{\cos \alpha}\right)}
$$
The answer is $V>\sqrt{g\left(2 H+l \frac{1-\sin \alpha}{\cos \alpha}\right)}$