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Calculate the pressure of a fluid at the bottom of the container

I have a container, the bottom of which is fixed to the floor. A fluid is poured into the container, then a time $T$ is allowed for the fluid to drain. The volume of the liquid is $V=11\,\mathrm{cm^3}$. The surface area of the container’s bottom is $A = 12\,\mathrm{cm^2}$. Find the maximum pressure and the speed at which the fluid is drained when $T = 0.5\,\mathrm{s}$.

So I know that to compute the maximum pressure I would have to find the maximum internal pressure and find the normal force acting on the bottom of the container but there isn’t much info on how to approach the problem. Since I’m really only familiar with solving problems involving pressure, I’m not really sure how to go about this one.

A:

You need the equivalent weight of the water to find the maximum pressure.
$$P_{top} = \frac{f}{A} = \frac{50\cdot g}{12\cdot12} = 3.8\cdot10^5 \ \mathrm{Pa}$$
You’d then find the flow rate as:
$$Q = \frac{V}{T} = \frac{11\cdot3.8\cdot10^5\cdot10^{ -6}}{0.5\cdot10^{ -2}} \approx 2.66\cdot10^{ -7}\ \mathrm{m^3s^{ -1}}$$
And speed as:
$$v = \frac{Q}{A} = \frac{2.66\cdot10^{ -7}\cdot12\cdot12}{
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