I'm working through an old UK A-Level Applied Mathematics textbook and am encountering a few problems. Here is one I'm struggling with - the second part of the question only.

Two identical elastic strings of length 1 metre and modulus 4.9 N are each fastened to a particle of mass 0.5 kg. Their other ends are fixed to two points 4 metres apart in a vertical line. Find the height of the particle above the lower fixed point A in the equilibrium position. The particle is now pulled down to A and released from rest. Find the greatest height above A to which the particle rises.

For the first part, here is my answer:

Let the tensions in the strings beT_{1},T_{2}and let the extension in the upper string bex.

The particle is in equilibrium, soT_{1}-T_{2}= 0.5g = 4.9

Also, by Hooke's Law,T_{1}= 4.9xandT_{2}= 4.9(2 -x)

Therefore 4.9x- 4.9(2 -x) = 4.9 sox= 1.5 m.

So, the height of the particle above A in equilibrium is 3 -x= 1.5 m.

Here is my attempt at the second part, using Conservation of Mechanical Energy. Although I get the same numerical answer as the book, they have it as 2√2 m, and I don't understand how they get that, so maybe I'm doing something wrong.

Lethbe the greatest height above A reached by the particle. Then considering the mechanical energy of the system:

At the initial position: Elastic Potential Energy = 4.9 x 3^2/2 = 22.05 J

When the particle reaches its highest point: Potential Energy = 0.5gh= 4.9h(taking g = 9.8 m s^{-2})

Elastic Potential Energy

(because the lower string is pulling the particle down, whilst the upper string is pulling it up)

Total mechanical energy is conserved, so 4.9h+ 9.8(h- 2) = 22.05

i.e.h= 17/6 = 2.83 m

Now, to 2 dp this is the same numerical value as the solution of 2√2 m given in the book, but I'd really like to know how the authors got to that solution. What am I doing wrong?