When a cable is suspended from two points, it hangs in the form of a catenary, the equation of the curve being given by 
When a cable is suspended from two points, it hangs in the form of a caternary, the equation of the curve being given by
  $$
s = 2a \sinh{\frac{x}{a}}\tag{1}
$$
$$
y = a\left( \cosh{\frac{x}{a}}-1 \right)\tag{2}
$$
  where $y$ is the sag of the cable, $x$ is the horizontal distance from the midpoint to one end of the cable, $a$ is the height of the lowest point of the cable and $s$ is the cable length.

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A heavy cable of 15 m is hanging between poles with the cable attached to each pole at the same height above ground level. 
  At the midpoint between the poles the cable should be 5.6 m above ground level. 
  Find
  
  
*
  
*i)  the required distance between the two poles
  
*ii) the minimum height of each pole.
  

 A: Thanks for replying to my comment. Unfortunately, it did not help me to clear my doubts. Therefore, I will use the standard method of solving your problem assuming what you have mentioned in your text is correct. 
That also means I am ignoring your sketch, which I think is wrong, because it dose not agree with the first equation. Therefore, I am attaching my own sketch for you to follow and understand the text given below. 

The length of the cable $s$ is equal to $\bf{15}$ m. Since the height of the lowest point of the cable is $\bf{5.6}$ m, $a=5.6$. According to the diagram, $x=FE$. From (1), we have,
$$s=2a\sinh\left(\frac{x}{a}\right) \space \to\space 15=2\times5.6\times\sinh\left(\frac{FE}{5.6}\right).$$
$$\therefore FE=5.6\times \sinh^{-1}\left(\frac{15}{2\times 5.6}\right)\approx6.1722\space \rm{m}$$
Now we use (2) to calculate the sag $y$, which is denoted by $DB$.
$$y=a\left(\cosh\left(\frac{FE}{a}\right)-1\right)\space\to\space DB=5.6\times\left(\cosh\left(\frac{6.1722}{5.6}\right)-1\right)=3.76\space\rm{m}$$
The distance between the two poles is $FG$ and it is equal to $2x$, i.e. 
$$FG=2\times FE=2\times 6.1722=12.3444\space \rm{m}.$$
The minimum height of each pole is $DE$ and is given by $\left(a+y\right)$, i.e.
$$AF=CG=DE=DB+a=3.76+5.6=9.36\space \rm{m}.$$
$\color{red} {\underline{Note}:}$
Because I could not agree with you on certain parts of your problem description, I cannot take responsibility for the values I gave above as answers to your question. Therefore, you use them at your own risk. At the same time, feel free to comment on my answer.
