Application of Torricelli’s theorem (8284)

, por F_y_Q

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A cylindrical container is filled with a liquid up to a height of one meter from the base of the container. Then, a hole is made at a point 80 cm below the liquid level:

a) What is the exit velocity of the liquid through the hole?

b) At what distance from the container will the first drop of liquid that touches the ground fall?

P.-S.

a) The exit velocity of the liquid through the hole is given by the expression:

\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}

Since the hole is made at a distance of 0.8 m from the liquid level:

v = \sqrt{2\cdot 9.8\ \frac{m}{s^2}\cdot 0.8\ m} = \fbox{\color[RGB]{192,0,0}{\bm{3.96\ \frac{m}{s}}}}


b) To calculate the distance at which the first drop falls, you must consider that it follows a motion similar to a horizontal launch. In this case, the position with respect to the X and Y axes follows the equations:

\left {\color[RGB]{2,112,20}{\bf x = v\cdot t}} \atop {\color[RGB]{2,112,20}{\bm{y = \frac{1}{2}gt^2}}}} \right \}

Since you know that the drop starts at a height of 0.2 m:

t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2\cdot 0.2\ \cancel{m}}{9.8\ \frac{\cancel{m}}{s^2}}}= \color[RGB]{0,112,192}{\bf 0.2\ s}

To find the horizontal position, substitute this time:

x = v\cdot t = 3.96\ \frac{m}{\cancel{s}}\cdot 0.2\ \cancel{s}= \fbox{\color[RGB]{192,0,0}{\bf 0.79\ m}}

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