4. (4 points) When moving in the air, a ball is acted upon by a resistance force proportional to the square of its velocity. Immediately before the volleyball player's hit, the ball was flying horizontally at a speed of $V_{1}$. After the hit, the ball flew vertically upwards at a speed of $V_{2}$ with an acceleration of $a_{2}$. Determine the acceleration of the ball immediately before the hit.

Possible solution. Let's write the equation connecting the ball's acceleration and its velocity in horizontal motion:

$$
a_{1}=\frac{F_{1}}{m}=\frac{k V_{1}^{2}}{m}
$$

Here $F_{1}$ is the resistance force, $m$ is the mass of the ball, and $k$ is the proportionality coefficient. After the hit, the ball flew upwards, and the forces of gravity and resistance acted on it, both directed downwards. Its acceleration in this case will be

$$
a_{2}=\frac{k V_{2}^{2}+m g}{m}=\frac{\frac{m a_{1}}{V_{1}^{2}} V_{2}^{2}+m g}{m}=a_{1}\left(\frac{V_{2}}{V_{1}}\right)^{2}+g
$$

For the acceleration $a_{1}$, we get

$$
a_{1}=\left(\frac{V_{1}}{V_{2}}\right)^{2}\left(a_{2}-g\right)
$$
Answer: $a_{1}=\left(\frac{V_{1}}{V_{2}}\right)^{2}\left(a_{2}-g\right)$.

## Evaluation Criteria

| Performance | Score |
| :--- | :---: |
| The participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| The expression for acceleration in horizontal flight is written | $\mathbf{1}$ |
| The expression for acceleration in vertical flight is written | $\mathbf{1}$ |
| Necessary transformations are performed and the answer is obtained | $\mathbf{2}$ |
| Total points | $\mathbf{4}$ |