It is raining heavily. Drop rate u. The ball slides on the asphalt at a speed of v. How many times in the same period of time does it get more drops than the same, but stationary ball? Does the answer change if the ball is not round?

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    <h3> Statement </h3>
    <p>
        $1.4.17.$ It is raining heavily. Drop rate $u$. The ball slides on the asphalt at a speed of $v$. How many times in the same period of time does it get more drops than the same, but stationary ball? Does the answer change if the ball is not round?
    </p>

    <h3>Solution</h3>
    <p>
        <p>During time $t$, all the drops that are in a cylinder of height $ut$ and base area $S_1$ equal to the area of ​​the central cross-section of the ball perpendicular to $u$ fall on a stationary ball. That is, in the first case</p>

$$N_1 = utS_1$$

In the second case, we move to the reference frame of a rolling ball, then the drops fall on it with a speed of

$$u_1 = \sqrt{u^2 + v^2}$$

$$N_2 = u_1 t S_2$$

where $S_2$ is equal to the area of ​​the central cross-section of the ball perpendicular to $u_1$. For a ball-ball $S_1 = S_2 = S$, and then

$$N_2 = \sqrt{u^2 + v^2} t S$$

Where we get the desired ratio

$$\fbox{$\dfrac{N_2}{N_1} = \sqrt{1 + \dfrac{v^2}{u^2}}$}$$

When the ball has a shape different from a sphere, its cross-sectional area will vary from the plane of the section

$$S_1\neq S_2\neq S$$

Thus, the answer obtained will also vary depending on the shape of the ball

NO: For additional reading, I recommend the discussion of this problem on ask.bc-pf.org

    <h4>Answer</h4>
    <p>
        <p>More drops fall on a rolling ball in $\dfrac{N_2}{N_1} = \sqrt{1 + \dfrac{v^2}{u^2}}$ times.</p>
    </p>



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      Aliaksandr Melnichenka<br>
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