- #1
Lotto
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- Homework Statement
- The barrel of a cylindrical shape containing the ##A_1## cross-section has a discharge hole in the bottom containing the ##A_2## cross-section. ##A_1## is much bigger than ##A_2##. Water's coming out through the opening. At time ##t_0 = 0##, the level has a height of ##h_0## above the bottom of the barrel.
Calculate the level's velocity as a function of time, its acceleration and the time when the barrel is empty.
- Relevant Equations
- ##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##
Here is only my solution:
##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##,
so by integrating we get
##h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.##
Setting ##h(T)=0## we get
##T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.##
By doing the first time derivative of ##h## we get
##v(t)=\frac{{A_2}^2}{{A_1}^2}gt-\frac{A_2}{A_1}\sqrt{2h_0 g}.##
##v(t)## has negative value because it points downward.
By doing the first time derivative of ##v## we get
##a(t)=\frac{{A_2}^2}{{A_1}^2}g.##
It is correct? Is there an "easier" way to solve it?
##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##,
so by integrating we get
##h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.##
Setting ##h(T)=0## we get
##T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.##
By doing the first time derivative of ##h## we get
##v(t)=\frac{{A_2}^2}{{A_1}^2}gt-\frac{A_2}{A_1}\sqrt{2h_0 g}.##
##v(t)## has negative value because it points downward.
By doing the first time derivative of ##v## we get
##a(t)=\frac{{A_2}^2}{{A_1}^2}g.##
It is correct? Is there an "easier" way to solve it?