Solving Diff Eq for Free-fall w/o Drag in Tank Outlet

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Suppose a tank containing a liquid has an outlet near the bottom. The velocity of the liquid flowing out of the outlet is equal to a particle free falling without drag.

a. show that v=sqrt(2gh)

b. by finding the rate of outflow to the rate of change of the liquid in the
tank, find h(t) that satisfies the equation:
A(h)(dh/dt) = (-b)(a)(sqrt(2gh))

A(h) = the area of the cross section of the tank at height h
b = a contraction coefficient that describes how the water flows (constant)
a = the radius of the outlet
g = gravity (9.81 m/s^2)

c. Consider a water tank in the form of a right circular cylinder. The water level is 3m above the outlet. Radius of tank is 1m radius of outlet is 0.1m. the value for the constant "b" is 0.6 How long does it take to drain the tank to the level of the outlet.

I have solved part a and really need help for the rest. If anyone can help it would be greatly appreciated. Also if you don't mind doing part A i would like to check my method.
 
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Hi and welcome to Physicsforums he lives.
Before we help, we need to see some effort on your side. Can you post your attempt at this problem, along with your thoughts or ideas on how to solve it?
 
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