A projectile must hit a target that is $A=40 \mathrm{~m}$ away and $B=30 \mathrm{~m}$ high, with the smallest possible initial velocity. At what angle should the launch occur?
I. solution. The coordinates of an object thrown at an angle $\alpha$ with an initial velocity $c$ are:

$$
\begin{aligned}
& x=c \cdot \cos \alpha \cdot t \\
& y=c \cdot \sin \alpha \cdot t-\frac{g}{2} \cdot t^{2}
\end{aligned}
$$

Eliminating time, we get the trajectory function:

$$
y=\operatorname{tg} \alpha \cdot x-\frac{g}{2 c^{2} \cos ^{2} \alpha} \cdot x^{2}=\operatorname{tg} \alpha \cdot x-\frac{g}{2 c^{2}} \cdot\left(1+\operatorname{tg}^{2} \alpha\right) x^{2}
$$

In our problem, $x=A$ and $y=B$ are given values. Substituting these into the trajectory function, we get an equation for the initial velocity $c$:

$$
B=\operatorname{tg} \alpha \cdot A-\frac{g}{2 c^{2}} \cdot\left(1+\operatorname{tg}^{2} \alpha\right) A^{2}
$$

From this, the initial velocity is:

$$
c^{2}=\frac{g A}{2} \cdot \frac{1+\operatorname{tg}^{2} \alpha}{\operatorname{tg} \alpha-B / A}
$$

The derivative of the variable part is:

$$
\frac{2}{g A} \cdot \frac{d\left(c^{2}\right)}{d(\operatorname{tg} \alpha)}=\frac{\operatorname{tg}^{2} \alpha-2 B \operatorname{tg} \alpha / A-1}{(\operatorname{tg} \alpha-B / A)^{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_bcfdc3109d2ef5a457b6g-1.jpg?height=488&width=338&top_left_y=1035&top_left_x=880)

Figure 1

A critical point is possible at the value of $\operatorname{tg} \alpha$ where the derivative is zero:

$$
\operatorname{tg}^{2} \alpha-2 B \operatorname{tg} \alpha / A-1=0
$$

The solution to this is:

$$
\operatorname{tg} \alpha=\frac{B}{A}+\frac{\sqrt{A^{2}+B^{2}}}{A}
$$

In our result, $C=\sqrt{A^{2}+B^{2}}$ represents the distance from the origin to the target point (Figure 1). Using $C$:

$$
\operatorname{tg} \alpha=\frac{B+C}{A}
$$

We get the initial velocity by substituting the value of $\operatorname{tg} \alpha$ into (1):

$$
c^{2}=\frac{g A}{2} \cdot \frac{1+(B+C)^{2} / A^{2}}{(B+C) / A-B / A}=g(C+B)
$$

With our data, $\operatorname{tg} \alpha=(30+50): 40=2, \quad \alpha=63.4^{\circ} ; \quad c=\sqrt{g(B+C)}=28 \mathrm{~m} / \mathrm{s}$.

George Bérces (Nagykanizsa, Landler Gym., IV. grade)

The result can also be obtained by construction. Measure the distance $C=O P=P Q$ vertically above the target point $P$ and draw the line $O Q$. The initial velocity $c$ is equal to the velocity with which an object can be thrown vertically upward to a height of $(B+C)/2$.

Julia Faragó (Bp., Jedlik Á. Gym., IV. grade) and

Peter Véner (Bp., Kaffka M. Gym., IV. grade)

II. solution. It is instructive to solve the problem geometrically. (See the article on page 33 of the 6th issue of KML in 1962.)

The farthest points that can be reached with the given initial velocity lie on the envelope of the projectile trajectories (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_bcfdc3109d2ef5a457b6g-2.jpg?height=542&width=371&top_left_y=136&top_left_x=866)

Our target point $P$ is also on this envelope, and we need to find the envelope knowing this point. It is known about the envelope that its focus is at the origin, and its directrix is at a height of $C=O P=P Q$. Thus, the value of $h=c^{2} / 2 g$ characteristic of vertical projection is $(B+C) / 2$,

$$
\frac{c^{2}}{2 g}=\frac{B+C}{2}
$$

from which it follows that the initial velocity is $c=\sqrt{g(B+C)}$.

The directrix of the projectile parabola is the horizontal line at a height of $O Y=h$. The geometric locus of the focus is the circle with center $O$ and radius $h$. Point $P$ is on the envelope, and thus can be found on only one trajectory, but it is also a point on the projectile parabola. The focus of the projectile parabola is the intersection point $F$ of the line $O P$ and the circle with radius $O Y=h$. To find the launch angle $\alpha$ corresponding to the minimum velocity, we only need to bisect the angle between the direction $O P$ and the $y$-axis.