
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

In a reference frame associated with Gavrila, balls are thrown upwards from a single point with the same speed at intervals of $\tau$ time. Introducing the $y$-axis directed vertically upwards with its origin at the point of throw and measuring time from the moment the first ball is thrown, the motion equations for both balls are:

$$
y_{1}(t)=U t-\frac{g t^{2}}{2}, \quad y_{2}(t)=U(t-\tau)-\frac{g(t-\tau)^{2}}{2}
$$

The latter equation is relevant for $t \ge \tau$. The distance between the balls is given by $s(t)=\left|y_{1}(t)-y_{2}(t)\right|=\mid U \tau-g t \tau + \frac{g \tau^{2}}{2} \right\rvert\,$. We need to find the minimum of this function.

If $2 U \ge g \tau$, there is a value $T=\frac{\tau}{2}+\frac{U}{g} \ge \tau$ at which $s(T)=0$, which is obviously the smallest distance between the balls. Assuming that the interval $\tau$ is small, let us consider this inequality as satisfied.

In the considered reference frame, Glafira is moving away from the throwing site and the line along which the balls fly at a speed of $V$, so by the time the balls meet, the distance between her and the balls will be $V T=V\left(\frac{\tau}{2}+\frac{U}{g}\right)$.

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