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Application of Torricelli’s theorem (8284)
Lunes 12 de agosto de 2024, por
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?
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}}}](local/cache-vignettes/L141xH26/6b8ef53b9924732e6103f56f9f68c5d6-f9d1b.png?1732980468 "\color[RGB]{2,112,20}{\bm{v= \sqrt{2\cdot g\cdot h}}}"){kind=small}
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}}}}](local/cache-vignettes/L324xH52/4255c299f7ff628166fc0ce03791540a-f0379.png?1732980468 "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 \}](local/cache-vignettes/L105xH54/031243fec7947d73e589855882668bd5-5c815.png?1732980468 "\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}](local/cache-vignettes/L291xH65/8cc8244ea60608b15a7b13fca5849665-f731b.png?1732980468 "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}}](local/cache-vignettes/L343xH45/43a81991bb4b60becdd55b81eb5a1c8b-32f74.png?1732980468 "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}}"){kind=small}