You will explore the characteristics of ballistic motion in cases where a body collides with and is reflected elastically from massive obstacles during flight. Air resistance can be neglected in this problem. Acceleration due to gravity is $g=9.8~{\rm m/s}^2$.
The following mathematical facts may be useful. Consider the polynomial of the second degree $f(x)=ax^2+bx+c$, $a<0$.
Consider the collision of a body with a massive vertical wall. We will work in a Cartesian coordinate system such that the $Oy$ axis is vertical. Let the velocity of the body in the instant before the collision be $\vec{v}_0=(v_{0x}, v_{0y})$. The wall moves with velocity $\vec{u}=(-u,0)$, $u>0$.
Since the obstacle is massive and the impact is perfectly elastic, in the frame of reference fixed to the wall, the magnitudes of the body's velocity before and after the collision are identical, and the angle of reflection is equal to the angle of incidence. This assumption must be maintained throughout the problem.
Now let's assume that the $Oy$ axis is directed vertically upwards, and the $Ox$ axis is horizontal and lies in the plane of the throw. The body is located at the origin of the coordinate system $x=y=0$. At the initial moment of time ($t=0$), the body is thrown with initial velocity $\vec{v_{0}}=(v_{0x}, v_{0y})$.
It is known that at times $t_1$ and $t_2$ after the throw, the body was at the same height above the earth's surface.
There is a massive vertical wall at the point where the body in question should land. Simultaneously with the launch of the body, it begins to move towards it.
Two experiments were conducted, as described in the problem statement. Their difference lay solely in the initial velocity of the wall. It is known that in the first case, the collision occurred at time $t_1$, and in the second - at time $t_2$, but in both cases at the same height. The distance between the places where the body falls in the first and second experiments is equal to $L$.
Let us now consider the launch of the body from the starting point with a certain fixed initial velocity $v_0$, but a free launch angle $\alpha$. A safety parabola is the boundary of an area beyond which a projectile cannot hit, regardless of the angle of launch $\alpha$. First you need to get the equation of the safety parabola.
Let again axis $Oy$ is vertical, axis $Ox$ is horizontal and lies in the plane of the throw. The body is located at the origin of the coordinate system $x=y=0$.
Assuming that the boundary of the region is a parabola, we can conclude that its equation has the form $y=ax^2+bx+c$.
The boy throws a small ball from a height of $H$. The initial velocity is $v_0$.
Now let the person perform the same throw, being strictly in the middle between two vertical walls, the distance between which is $l$.
Hint. Imagine that there is a mirror hanging on the wall against which the body is hitting. What will be the movement of the body after the bounce in the mirror?