38.6. Find the envelope of the trajectories of a material point thrown from the origin with a velocity $v_{0}$ in a fixed vertical plane.

\ The curve from problem 38.6 is called the safety parabola.
38.6. Let's direct the $O y$ axis vertically upwards, and the $O x$ axis along the horizontal component of the velocity $v_{0}$. Then at time $t$, the material point has coordinates $x(t)=v_{0} \cos \alpha \cdot t$, $y(t)=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}$, where $\alpha$ is the angle at which it was launched. For example, if the point is launched vertically upwards, then $y(t)=v_{0} t-\frac{g t^{2}}{2}$, so $y(t)$ is maximal at $t=\frac{v_{0}}{g}$ and at this time $y(t)=\frac{v_{0}^{2}}{2 g}$. If we assume that the envelope is a parabola, then it should be given by the equation $y=\frac{v_{0}^{2}}{2 g}-k x^{2}$. To find $k$, we will calculate the greatest possible coordinate of the intersection of the trajectory with the $O x$ axis (the greatest range of the shot). If $y\left(t_{0}\right)=0$ and $t_{0} \neq 0$, then $t_{0}=\frac{2 v_{0} \sin \alpha}{g}$. At this time, $x\left(t_{0}\right)=\frac{2 v_{0}^{2} \sin \alpha \cos \alpha}{g}=\frac{v_{0}^{2} \sin 2 \alpha}{g}$. Therefore, the greatest range of the shot is $\frac{v_{0}^{2}}{g}$. Accordingly, for $k$ we get the equation $y=\frac{v_{0}^{2}}{2 g}-k\left(\frac{v_{0}^{2}}{g}\right)^{2}=0$, from which we find $k=\frac{g}{2 v_{0}^{2}}$.

Now let's prove that the parabola $y=\frac{v_{0}^{2}}{2 g}-\frac{g}{2 v_{0}^{2}} x^{2}$ is indeed the envelope of the considered family of trajectories. For this, it is sufficient to prove that any trajectory lies below this parabola and has one common point with it, i.e., for all $t$ the inequality

$$
\frac{v_{0}^{2}}{2 g}-\frac{g \cos ^{2} \alpha}{2} t^{2} \geqslant v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}
$$

holds, and for some $t$ it turns into an equality. Replacing $\cos ^{2} \alpha$ with $1-\sin ^{2} \alpha$, we get the inequality

$$
\frac{v_{0}^{2}}{2 g}-v_{0} \sin \alpha \cdot t+\frac{g \sin ^{2} \alpha}{2} t^{2} \geqslant 0
$$

i.e., $\frac{g}{2}\left(\frac{v_{0}}{g}-\sin \alpha \cdot t\right)^{2} \geqslant 0$. It turns into an equality at $t=\frac{v_{0}}{g \sin \alpha}$.