5. (10 points) Two stones are thrown simultaneously with the same initial speed $v_{0}$. The first stone is thrown horizontally from a height of $H=50$ m, and the second is thrown vertically upward from the surface of the Earth. It is known that the stones collided in the air. Determine the initial distance between the stones. Neglect air resistance.
Answer: $\approx 70.7$ m.

Solution. Equations of motion for the first stone: $y_{1}=H-\frac{g t^{2}}{2}$.

$x_{1}=v_{0} t=L$,

where $L$ is the initial horizontal distance between the bodies.

Equation of motion for the second body: $y_{2}=v_{0} t-\frac{g t^{2}}{2}$.

Since the stones met, $y_{1}=y_{2}$,

As a result: $H=v_{0} t=L$.

We obtain that the initial distance between the stones:

$S=\sqrt{2} H=\sqrt{2} \cdot 50 \approx 70.7 \mathrm{m}$.