- #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?