There are two principles at work here: 1) Bernoulli's Principle, which states that
$$2 g y + v(y)^2 = v(0)^2$$
where $v=dy/dt = \dot{y}$ is the rate at which the fluid is sinking, and $y$ is the height of the fuid. Also, $g$ is the acceleration due to gravity $\approx 9.8 \text{m}/\text{sec}^2$. $v(0)$ is the speed of the fluid exiting the hole at the bottom.
2) $A(y) v(y) = B v(0)$ - this is a statement that the amount of fluid exiting the container is uniform throughout - sort of a conservation principle. This allows us to get a diffrential equation for the height of the fluid at any time:
$$\dot{y}^2 = \frac{2 g y}{\pi^2 y^2/B^2 - 1}$$
Note that I used the fact that the area of the fluid $A(y)$ at height $y$ is $\pi y$ for this container.
In principle, we have a simple ODE which may be expressed in terms of an integral over $y$; the integral, however, is pretty horrid (I get elliptic integrals over imaginary arguments). Nevertheless, we may exploit the fact that the area of the hole at the bottom is very small so that the area of the fluid above the hole at any time is much greater than $B$. We may therefore neglect the $1$ in the denominator and get the approximate DE:
$$\dot{y} = \pm \frac{B}{\pi}\sqrt{\frac{2 g}{y}}$$
We choose the negative sign because $y$ is decreasing. The solution to this equation takes the form
$$y^{3/2} = y(0)^{3/2} - \frac{3 B}{2 \pi} \sqrt{2 g} t$$
You may then find the approximate time at which the container is empty by setting the above to zero.
Hint: We know that the surface area of the cone, $A= \Pi r^2$ and the volume, $V = \frac{1}{3}\Pi r^2x$. Since $V$ is given in the question, we can find that $r^2 =\frac{512}{x}$ and substitute the $r^2$ in the area equation and write it in terms of $x$, which gives us $A(x)=\frac{512\Pi}{x}$.
Lastly, you use $\frac{dx}{dt}$ on Maple to find the solution. Look into p.290 of Burden and Faires's textbook Numerical Analysis (ninth edition) for an idea on how to write the Maple code. I assume you're using this text b/c the question you asked is right from that very textbook.
Best Answer
I'm doing the exercise too.
Questions 1 is simple. Denote the area of orifice $A_0$ and we'll have $ \frac{dV}{dt}=- c\sqrt{2gy}A_0$ and $\frac{dV}{dy}=A(y)=4$.use the chain rule we have $$\frac{dy}{dt}=\frac{dy}{dV} \frac{dV}{dt} = -\frac{c \sqrt{2gy}A_0}{A(y)}$$ cause $c\sqrt{2g}A_0 /A(y)$ is constant ,denote it by $K$ and rewrite the formula we get $$\frac{y'}{\sqrt{y}}=-K$$Integrate it with the data given $$\int_2^1 \frac{y'}{\sqrt{y}}=-K \int_0^xdt$$ .The only thing have to take notice is the unit of $A_0$ is "square inches" but the one of A(y) is "square feet".
Question 2. I get a idea to solve it.
Denote the volume of water flowed in by $Vi$ ,we have $\frac{dV}{dt}=V_i-c\sqrt{2gy}A_0$,$\frac{dV}{dy}=A_y$ ,
use the chain rule to get $$\frac{dy}{dt}=\frac{dy}{dV} \frac{dV}{dt} = \frac{V_i - c \sqrt{2gy}A_0}{A_y}$$ let's say $y'=n-m\sqrt{y}$ ($n$ and $m$ are both positive)for simplicity. And let $r= 1/2$ and $$g=y^{1-r}=\sqrt{y},g'=(1-r)y^{-r}y'$$ we have
$$g'=(1-r)(ng^{-1}-m)=\frac{1}{2}\frac{n-mg}{g}$$ it's seperatable.Rewrite it and get$$(\frac{1}{m}+\frac{n}{m^2} \frac{m}{mg-n})g' = - \frac{1}{2}$$ Solve this equation we have an implicity formula:$$\frac{g}{m} + \frac{n}{m^2}\log|mg-n|=-\frac{t}{2}+C (or) \frac{\sqrt{y}}{m} + \frac{n}{m^2}\log|m\sqrt{y}-n|=-\frac{t}{2}+C$$ for some constant $C$.
The question is $y \to (\frac{25}{24})^2$ as $t \to + \infty$,notice the RHS approach to $-\infty$,so the LHS should approach to $-\infty$ too. Because the first term of the LHS is always opsitive.The only way to achieve this is $\log|m\sqrt{y}-n| \to - \infty$ means $m\sqrt{y}-n \to 0 $ as $t \to + \infty$ ,so $y$ approaches to $(\frac{n}{m})^2=(\frac{25}{24})^2$ as $t \to + \infty$. Check the answer we'll see $$K + -\infty = -\infty + C$$