Why Is v1 Assumed to Be Zero in Bernoulli's Equation for a Large Tank?

[TheCurator]

Homework Statement

Q. A large water tank, open at the top, is situated at the top of a hill. A pipe leads downhill from
the bottom of the tank. The pipe has a diameter 5.0 cm where it is attached to the tank at point
A, but gradually narrows to a diameter of 3.0 cm at a point B which is 12.0 m below the
surface level of the water in the tank.
Velocity at A is 4.0 ms.

(ii) Calculate the absolute pressure at the lower level (point B).

Homework Equations

Values:
p1 = pO = 1.01 * 10^5
P(density) = 1.00 * 10^3
v1 = 4.0 (back to this in a second)
y1 = 0 (assume, since the question indicates nothing else)

p2 = ?
v2 = (v1A1)/A2 = 11 ms
y2 = 12 m

Rearrange bernoullis;

p2 = p1 + 1/2P(v1^2 - v2^2) + Pg(y1 - y2)
Sub in, calculate to get answer.

The answer given in the exam booklet gives v1 as 0, while y1 = 12 and y2 = 0.
I'm pretty sure the y values were swapped because it essentially achieves the same purpose, ie. the distance between the two points.
However, I have no idea as why v1 would be 0. Any help?
The booklet also states that we assume the tank to be large, so v1 ~ 0. Why can't we use the v1 = 4 value, however?

 

[Doc Al]

The book is comparing a point at the water surface (not at point A) to point B. You only have partial information about point A, so it's less useful. (What's the height of point A? The pressure?)

 

[TheCurator]

Thanks!