$1.3.27^*.$ 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?
The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in 1.3.6:
$$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$ $$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$
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 $v_0$
$$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$
$$v = \sqrt{5gR}$$