Hole in the bottom of the can of \( A_2 \) area.

Find time it takes to drain the can.

Yadda yadda Bernouli's Eq applies.

V = velocity, P = pressure, z = height, Row = density.

\(

P_1 + 1/2RowV^2 + Row*g*z = P_2 + 1/2RowV_2^2 + Row*g*z_2

\)

Pressures are equal outside the can, Row cancels, V_1 is negligible and \( z_2 \) = 0.

\(

gz=1/2V_2^2

V_2 = sqrt{2gz}

\)

Relationship between volume out of the can and change in height, feel like I'm wrong here:

\(

A_1 H/t=A_2 V_2

H = A_2 /A_1 *V_2 t

dh/dt = A_2 /A_2 V_2

\)

Regardless of whether that is right, I think this is my differential equation:

\(

dh/dt = -A_2 /A_1 V_2

dh/dt = -A_2 /A_1 sqrt{2gh}

\)

Separate Variables

\(

dh/sqrt{h}=-A_2 /A_1 sqrt{2g}dt

\int 1/sqrt{h} dh = \int -A_2 /A_1 sqrt{2g}dt

2sqrt{h} = -A_2/A_1 sqrt{2g} t +C

h = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2

\)

Solve for t when h = 0.

\(

0 = ((-A_2/A_1 sqrt{2g} t)/2 +C)^2

2C*A_1/A_2 * 1/sqrt{2g} = t

\)

Doesn't feel right, and my calculator is graphing it with an absurdly long time until empty.