Question

A farmer sees that a circular trough has a hole in the base. The hole is 3.0m below the level of the water in the tank. If the top of the trough is open to the atmosphere what is the speed of the water as it leaves the hole.

Answer 1

Let's write the Bernoulli equation (indexes 1 and 2 corresponds to the top and bottom, respectively):

$$\dfrac{\rho v_1^2}{2}+\rho gh_1+p_1=\dfrac{\rho v_2^2}{2}+\rho gh_2+p_2.$$

Since $v_1=0$, $h_2=0$ and $p_1=p_2$, we get:

$$\rho gh_1=\dfrac{\rho v_2^2}{2},$$ $$v_2=\sqrt{2gh_1}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot3.0\ m}=7.67\ \dfrac{m}{s}.$$