Calculus for the Practical Man: Chapter 4, Problem 16 
A rope 28 feet long is attached to a block on level ground and runs over a pulley 12 feet above the ground. The rope is stretched taut and the free end is drawn directly away from the block and pulley at the rate of 13 ft. per sec. How fast will the block be moving when it is 5 feet away from the point directly below the pulley?
My solution: 
I start with relating the three sides of the triangle, holding one side constant at 12 feet. My aim is to find the rate of movement of the rope, that is, the rate of change of the hypotenuse, h, which determines, and is therefore identical to, the rate of change of the position of the box.
$$h^2 = x^2 + y^2$$
$$h = \sqrt{x^2 + 144},$$
and since the rate of movement of the rope dictates the rate of movement of the box, 
$$\frac{dh}{dt} = \frac{x}{\sqrt{x^2 + 144}} \cdot \frac{dx}{dt}$$
Given that $h = 28 - 5 = 23$ feet when the box is 5 feet from the pulley, and that $$ x = \sqrt{h^2 - y^2} = \sqrt{23^2 - 12^2} = \sqrt{385},$$
it follows that
$$\frac{dh}{dt} = \frac{\sqrt{385}}{\sqrt{385 + 144}} \cdot 13 \approx 11.1 \textrm{ ft/s}.$$
The book's answer is $8 \textrm{ ft^2/s}.$
Did I do something wrong? My answer is so close to the book's.
 A: 
Given that $x$ and $y$ are functions of time, the following relationship should always hold true: $x^2+12^2=(16+y)^2$. Thus:
$$
2x\frac{dx}{dt}+0=2(16+y)\frac{dy}{dt}\implies\\
\frac{dy}{dt}=\frac{x}{16+y}\cdot\frac{dx}{dt}.
$$
When the block is $5\ \text{ft}$ away from the point directly below the pulley, $y=7\ \text{ft}$. $\frac{dx}{dt}$ is known. It's $13\ \text{ft/sec}$. What is $x$ at the time when the block is $5\ \text{ft}$ away from the point directly below the pulley? It should be the positive solution to this equation:
$$
x^2+12^2=(16+7)^2\implies\\
x=\sqrt{385}\ \text{ft}.
$$
Therefore:
$$
\frac{dy}{dt}=\frac{\sqrt{385}}{16+7}\cdot 13\approx 11.1\ \text{ft/sec}.
$$
I don't know, but that's the answer I get. I don't see anything wrong with my logic.
A: $h^2  = x^2 + y^2$
$y = 12$ as identified.  
When $x = 5, h = 13$
$h^2 = 5^2 + 12^2 = 13^2$
That the length of the rope is 28' seems to be a red herring, except insofar as it establishes an upper bound for $x.$  We really only care about the length of rope from the pully to the mass.  The rope from the pully to the free end is irrelevant.
When we differentiate:
$2h\frac {dh}{dt} = 2x \frac {dx}{dt}$
$\frac {dh}{dt} = 13$
$2(13)(13) = 2(5)\frac{dx}{dt}\\
x = \frac{13^3}{5}$ 
