It is difficult only because it is in a word problem form.

Many are afraid of word problems.

dh/dt = -c*(A_(h)/A_(w))*sqrt(2*g*h)

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

I assume g = 32 ft/sec/sec. (It's supposed to be 32.2 ft/sec/sec.)

dh/dt = -c(1/4 /50)sqrt(2*32*h)

dh/dt = -c(1/200)(8sqrt(h)

dh/dt = -c(0.04)sqrt(h)

dh/sqrt(h) = -0.04c*dt

h^(-1/2) *dh = -0.04c *dt

Integrate both sides, the c is a constant,

2sqrt(h) = (-0.04c)t +C ------------------(i)

Given: when t=0, h=20

So,

2sqrt(20) = 0 +C

C = 4sqrt(5) ------------***

Substitute that into (i),

2sqrt(h) = (-0.04c)t +4sqrt(5)

sqrt(h) = (-0.02c)t +2sqrt5

Square both sides,

h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20 -----the equation.

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

When tank is empty, then h=0.

h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20

0 = (0.0004*1^2)t^2 -0.08sqrt(5)*1*t +20

0 = 0.0004t^2 -0.08sqrt(5)*t +20

Using the Quadratic formula,

(the discriminant is zero. Umm...)

t = 0.08sqrt(5) / 0.0008

t = 223.6 seconds ------------the tank is empty.

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

Again, when tank is empty, then h=0.

h = (0.0004c^2)t^2 -0.08sqrt(5)*c*t +20

0 = (0.0004*0.6^2)t^2 -0.08sqrt(5)*0.6*t +20

0 = 0.000144t^2 -0.048sqrt(5)*t +20

Using the Quadratic formula,

(the discriminant is zero again. )

t = 0.048sqrt(5) / 0.000288

t = 372.7 seconds ------------the tank is empty.