Equation of continuity for stones I used the same argument in the proof of equation of continuity to flow of stones. Suppose I drop stones from the upper end of a vertical pipe. I am continuously dropping the stones so that at any instant the pipe is full of stones. The stones clearly have streamlines just like fluid flow because they have only straight line downward motion. So, after applying law of conservation of mass, we get the equation of continuity for the stones. Since the stones are freely falling, so they clearly come out of the pipe with a speed greater than the initial speed. So, the  area of lower end of the pipe should be smaller. But this doesn't make sense because all the stones only have downward acceleration throughout their motion. So, there's no way their paths would have curved in the journey so that they come out of a smaller area. So, how's this possible? I've one confusion with the equation of continuity for fluid flow also. If I drop water from a height $(h)$ with initial velocity zero and having some finite initial area of tube. Then, the area of the tube after any finite time will be: $A_2=\frac{A_1v_1}{v_2} = 0.$ So, the water flow will have zero area after any finite time $(t).$ How's this possible?
 A: 1) About the rocks, you're ignoring the role of air. As they speed up, they separate, so air comes in to fill the spaces. With water, that can't happen (before drops form), so air pressure (and surface tension) squeezes the stream together.
2) About the water, you're ignoring the thickness $y_1$ of the slug of water, which has volume $A_1y_1$. After it has fallen and reached area $A_2$ it still has the same volume, so $y_2$ is larger.
A: The answer to the first question is that the rate of flow involves not just the velocity and cross sectional area, but also density.  With stones in a pipe, the area stays the same, so when the velocity rises, the density falls to compensate (the distance between stones get stretched vertically as their velocity rises).  With a fluid, we often assume interparticle forces that maintain a fixed density, so as the velocity rises, the cross sectional area is what drops rather than the density.  The interparticle forces, absent with stones, make that possible.
The answer to the second question is that the mass continuity equation assumes you have a steady state, i.e., the state of the fluid as a whole looks the same from moment to moment.  You can't have that if you want to keep feeding in new fluid at the top yet claim it is dropped with zero speed-- that fluid has to come from somewhere, so must have some kind of motion.  In the case of a water faucet, there is horizontal flow that is turning into vertical flow, so just analyze the flow along the pipe direction as the pipe turns from horizontal to vertical.  There is never anywhere that has zero speed along the pipe once a steady state appears.  And if you only look at the region where the flow is entirely vertical, there is always some nonzero vertical motion, even at the top of the downward flow.
A: First question:
The equation of continuity is:
d1*v1*a1=d2*v2*a2      where d1 and d2 are intial and final density, v1 and v2 are initial and final velocity and a1 and a2 is initial and final cross sectional area. The equation you are using is a special case of the above mentioned equation where the density remains constant. When we throw stones, their density is reduced. This is because the distance between them increases due to free fall. So the cross sectional area is not reduced and the equation is not violated as density is reduced.
Second question: The initial velocity in the equation is the rate at which the fluid enters the tube, initial velocity is zero implies that the fluid is not entering the tube. Try to understand that the initial velocity here cannot be zero because there has to be some rate at which the fluid is entering the tube. 
A: First of all, the 1st question has already been correctly answered by Ken G and Harmohit. This is for the comment you left on Ken G's answer and kind of related to your 2nd question. (I wanted to write this in the comments, but it was too long.) Anyway.
In the case of bucket, using $A_1 v_1=A_2 v_2$ and putting $v_1=0$ tells us that $v_2=0$ not $A_2=0$(as you have proposed) because we know that all the water is inside the bucket initially. So, $A_1=A_2$ (as bucket has a constant area). What it tells us is that no water has moved yet. In fact $A_1=A_2$ will be true till the time all the water is inside the bucket, and that will give you that $v_1=v_2$. And this also holds true when you look at the facts that we assume like, water moves as whole and with the same speed and also that it is always in contact with the pipe/bucket(in this case) until it is out of the pipe/bucket; after it is out, it will definitely change area, as now $A_1\not=A_2$ any more because water is free from the condition of always touching the pipe. Once it is out of bucket using $A_1 v_1=A_2 v_2$ and putting the values $v_1,v_2$ and $A_1$ will tell you the $A_2$. 
I know this will be difficult to understand because this is one of those things which are understood much better face to face.
