Solution

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

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

Substituting into the formula for $v_0$

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