Rain barrel

Relevant wiki: Bernoulli's Principle (Fluids)

The speeds of efflux $v_{0i}$ of the water jets results from energy conservation (Bernoulli's principle, Torricelli's law) $\begin{aligned} p_0 + \rho g h_i &= p_0 + \frac{1}{2} \rho v_{0i}^2 \\ \Rightarrow \quad v_{0i} &= \sqrt{2 g h_i}, \quad i = 1,2 \end{aligned}$ where $h_1 = h - \Delta h$ and $h_2 = h$ are the vertical distances of the holes to the water surface. The trajectories of the water jets are throwing parabolas. In horizontal $x$ -direction, they have a constant velocity $v_{0i}$ , while in vertical $z$ -direction there is a constant accelaration $g$ : $\begin{aligned} \left( \begin{array}{c} x_i(t) \\ z_i(t) \end{array}\right) &= \left( \begin{array}{c} v_{0i} t \\ z_{0i} - \frac{1}{2} g t^2 \end{array}\right) \\ \Rightarrow \quad z_i(x) &= z_{0i} - \frac{g}{2} \left(\frac{x}{v_{0i}} \right)^2 \\ &= -h_i - \frac{1}{4 h_i} x^2 \end{aligned}$ At a distance $x = s$ both parabolas intersect: $\begin{aligned} & & z_1(s) &= z_2(s) \\ \Rightarrow & & - (h - \Delta h) - \frac{1}{4 (h - \Delta h)} s^2 &= - h - \frac{1}{4 h} s^2\\ \Rightarrow & & \Delta h &= \frac{s^2}{4} \left( \frac{1}{(h - \Delta h)} - \frac{1}{h} \right) \\ \Rightarrow & & \Delta h &= \frac{s^2}{4} \frac{\Delta h}{(h - \Delta h)h} \\ \Rightarrow & & (h - \Delta h)h &= \frac{s^2}{4} \\ \Rightarrow & & h^2 - \Delta h \cdot h - \frac{s^2}{4} &= 0 \\ \Rightarrow & & h &= \frac{\Delta h + \sqrt{\Delta h^2 + s^2}}{2} \\ & & &= \frac{30 + \sqrt{30^2 + 40^2}}{2} \,\text{cm}\\ & & &= 40 \,\text{cm} \end{aligned}$