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**Ideasman** A water tank has the form of a right-circular cylinder. Through a whole in the bottom of the water tank, if water is able to pass through this whole with gravity acting upon it, the height, h, of water at time t, is given by the following Dif EQ:

dh/dt = -c*(A_(h)/A_(w))*sqrt(2*g*h) where A_(w) is the cross-sectional area of the water and A_(h) is the cross-sectional area of the hole. Further, c is the friction/contraction factor of the hole.

1.) Solve the Dif EQ if the initial height of the water is 20 ft and A_(w) = 50 ft^2 and A_(h) = 1/4 ft^2

2.) Suppose c = 1. Determine the time at which the tank is empty.

3.) If the friction/contraction factor, that is, if c = 0.6, how long would the tank take to empty?