Solution

First, let's determine the flight time of each water drop. This time is:

$$t = \frac{2v_0 \cdot \sin\alpha}{g}$$

There is water in the air, flowing out of the hose in exactly time $t$. The mass of this water is easy to determine: it is the mass of water in a cylinder of height $v_0 t$ and base area $S$.

$$m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin\alpha}{g}$$ $$m = \frac{2 \rho v_0^2 S \cdot \sin\alpha}{g} = 7.2 \text{ kg}$$