A can of height h and cross-sectional area A0 is initially full of water. A small hole of area A1 << A0 is cut in the bottom of the can.
Find an expression for the time it takes for all the water to drain out of the can.
Work:
Can has mass Mi and volume Vi at time t.
Can has mass (mi + dm) and volume (vi + dV) at a later time dt
h = height of can = k/A
Volume V = Ah
dV/dt = -k*sqrt(h), where sqrt = square root
Solution:
t = the integral of dt/dV