A spherical tank standing on the ground has a radius of R. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?

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    <h3 id="back-link"><a href="/#1.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.3.27^*.$ A spherical tank standing on the ground has a radius of $R$. What is the lowest speed at which a rock thrown from the ground can fly over the reservoir just by touching its top?
    </p>

    <h3>Solution</h3>
    <p>
        <p>The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in <a href="/1.3.6" target="_blank">1.3.6</a>:</p>

$$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$

The time it takes for the stone to rise to the maximum height $2R$ is found as

$$t_1 = \frac{v_0 \sin \alpha}{g}$$

The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore

$$\frac{v_0^2 \sin^2 \alpha}{2g} = 2R$$

Determine the value of the initial throw speed

$$v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}}$$

The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions

$$v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ $$2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$ $$\tan \alpha = 2; \quad \alpha = \text{arctg} \\;2 \approx 63^\circ$$

Substituting into the formula for $v_0$

$$\fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$}$$

    <h4>Answer</h4>
    <p>
        $$v = \sqrt{5gR}$$
    </p>


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