An Open Cylindrical Tank of Acid....

[The-Mad-Lisper]

Homework Statement

An open cylindrical tank of acid rests at the edge of a table $2.20\cdot 10^0\ m$ above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is $7.00\cdot10^{-1}\ m$ deep?
Let $\rho$ be the mass density of acid, $g$ be the acceleration of gravity, $h_{tank}$ be the depth of the tank, and $h_0$ the height of the base of the tank.

Homework Equations

$p_1+\rho g y_1+\frac{1}{2}\rho {v_1}^2=p_2+\rho g y_2+\frac{1}{2}\rho {v_2}^2$ (Bernoulli's Equation)

The Attempt at a Solution

$\Delta x = vt$

$\frac{1}{2}gt^2=h_0$
$t^2=\frac{2h_0}{g}$
$t=\sqrt{\frac{2h_0}{g}}$

$\rho g h_{tank}=\frac{1}{2}\rho v^2$
$gh_{tank}=\frac{1}{2}v^2$
$v^2=2gh_{tank}$
$v=\sqrt{2gh_{tank}}$

$\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}$
$\Delta x=\sqrt{\frac{2gh_{tank}h_0}{g}}$
$\Delta x=\sqrt{2h_{tank}h_0}$
$\Delta x=\sqrt{2\cdot 7.00\cdot 10^{-1}\ m\cdot 2.20\cdot 10^0\ m}$
$\Delta x=\sqrt{3.08\cdot 10^0\ m^2}$
$\Delta x = 1.75\cdot 10^0\ m$

Unfortunately, $1.75\ m$ is not the correct answer. Perhaps I am not setting up Bernoulli's equation correctly for this problem.

 

[SteamKing]

Homework Statement

An open cylindrical tank of acid rests at the edge of a table $2.20\cdot 10^0\ m$ above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is $7.00\cdot10^{-1}\ m$ deep?
Let $\rho$ be the mass density of acid, $g$ be the acceleration of gravity, $h_{tank}$ be the depth of the tank, and $h_0$ the height of the base of the tank.

Homework Equations

$p_1+\rho g y_1+\frac{1}{2}\rho {v_1}^2=p_2+\rho g y_2+\frac{1}{2}\rho {v_2}^2$ (Bernoulli's Equation)

The Attempt at a Solution

$\Delta x = vt$

$\frac{1}{2}gt^2=h_0$
$t^2=\frac{2h_0}{g}$
$t=\sqrt{\frac{2h_0}{g}}$

$\rho g h_{tank}=\frac{1}{2}\rho v^2$
$gh_{tank}=\frac{1}{2}v^2$
$v^2=2gh_{tank}$
$v=\sqrt{2gh_{tank}}$

$\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}$
$\Delta x=\sqrt{\frac{2gh_{tank}h_0}{g}}$
$\Delta x=\sqrt{2h_{tank}h_0}$
$\Delta x=\sqrt{2\cdot 7.00\cdot 10^{-1}\ m\cdot 2.20\cdot 10^0\ m}$
$\Delta x=\sqrt{3.08\cdot 10^0\ m^2}$
$\Delta x = 1.75\cdot 10^0\ m$

Unfortunately, $1.75\ m$ is not the correct answer. Perhaps I am not setting up Bernoulli's equation correctly for this problem.

No, the problem is not with the Bernoulli equation.

You have:

$t=\sqrt{\frac{2h_0}{g}}$

and

$v=\sqrt{2gh_{tank}}$

when you combine these two equations:

$\Delta x=\sqrt{gh_{tank}} \sqrt{\frac{2h_0}{g}}$

you dropped one of the factors of 2 under the radical sign for some reason.