Is this Image about boat resistance calculations correct?

Any rope will stretch to some extent depending on the load it is exposed to. Material with a high elastic modulus will stretch less for a given load but it will still stretch, and it will have some rebound if parted suddenly. I do not want to be in the path of 6x19 or any other rope, wire or otherwise, when it breaks because of overloading.

 

Yep.
BlueBell, think about it this way. Model the rope as two pieces of mass (m= m1+m2) that have length (L = l1+l2) and two spring (F=x1*k*(l2/L)= x2*k*(l1/L) where x is the stretch in each piece. The energy stored in each section of the rope is (xn^2*k)/2. When the rope breaks the mass of the rope is accelerated back by the stored energy available in the broken part, i.e. all potential energy is converted to kenitic energy v=sqrt(xn^2*k/mn). So if k is very large and mn is very light; the rope could come flying back with considerable velocity even if xn is rather small.