6. A body was thrown from the surface of the earth at some angle to the horizon. It is known that the velocity vector of the body is directed at an angle $\alpha=30^{\circ}$ to the horizon at times $t_{1}$ and $t_{2}$ after the throw. Find the maximum height of the body's rise above the ground and the distance from the point of throwing to the point of the body's fall to the ground. The acceleration due to gravity is $g$. Neglect air resistance.
6. Since the trajectory of the body is symmetric with respect to the highest point, the time of ascent $t_{\text {pod }}$ of the body to the maximum height lies exactly halfway between the moments $t_{1}$ and $t_{2}$:

$$
t_{\text {pod }}=\frac{t_{2}+t_{1}}{2}
$$

and the total time of motion $t_{\text {poli }}$ is twice the time (1)

$$
t_{\text {poli }}=t_{2}+t_{1}
$$

From formula (1) and the law of velocity change for uniformly accelerated motion, we find the change in the vertical component of the body's velocity over the time of ascent to the maximum height:

$$
0=v_{0, y}-\frac{g\left(t_{2}+t_{1}\right)}{2}
$$

where $v_{0, y}$ is the projection of the initial velocity vector onto the vertical axis. From this, we find the vertical component of the initial velocity of the body

$$
v_{0, y}=\frac{g\left(t_{2}+t_{1}\right)}{2}
$$

The horizontal component of the initial velocity can be found as follows. Since from the moment $t_{1}$ to the moment of ascent to the maximum height (i.e., over the time $\left.\left(t_{2}-t_{1}\right) / 2\right)$ the vertical component decreases to zero, applying the law of motion for velocity to this stage of motion, we get

$$
0=v_{1, y}-\frac{g\left(t_{2}-t_{1}\right)}{2}
$$

where $v_{1, y}$ is the vertical component of the velocity vector of the body at the moment $t_{1}$. Therefore,

$$
v_{1, y}=\frac{g\left(t_{2}-t_{1}\right)}{2}
$$

Therefore, the horizontal component of the velocity of the body at this moment is

$$
v_{1, x}=v_{1, y} \operatorname{ctg} \alpha=\frac{g\left(t_{2}-t_{1}\right)}{2} \sqrt{3}=v_{0, x}
$$

Since the horizontal component of the velocity does not change, it will be (3) throughout the motion. From here and formulas (2) we find the range of flight $S$

$$
S=v_{0, x_{\text {poli }}}=\frac{g\left(t_{2}-t_{1}\right)}{2} \sqrt{3}\left(t_{2}+t_{1}\right)=\frac{\sqrt{3}}{2} g\left(t_{2}^{2}-t_{1}^{2}\right)
$$

The maximum height of ascent $h$ can be found knowing the vertical component of the initial velocity

$$
h=\frac{v_{0, y}^{2}}{2 g}=\frac{g\left(t_{2}+t_{1}\right)^{2}}{8}
$$

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