$1.3.12^*.$ From a hose lying on the ground, water shoots at an angle of $45^\circ$ to the horizon with an initial velocity of $10$ $\frac{m}{s}$. The cross-sectional area of the hose opening is $5$ $cm^2$ . Determine the mass of the jet in the air.
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 $S$.
$$ m = \rho S v_0 t = \rho S \frac{2v_0^2 \cdot \sin \alpha}{g} $$ $$ m = \frac{2 \rho v_0^2 S \cdot \sin \alpha}{g} = 7.2 \text{ kg}. $$$$x = \frac{2\nu ^{2}}{g(tg\beta +tg\alpha )}$$