AB is a vertically suspended elastic cord of negligible mass and length L. Its force constant is $k=\frac{4mg}{L}.$ There is a massless  platform attached to the lower end of the cord. A monkey  of mass m starts from top end A and slides down  the cord with a uniform acceleration of  $\frac{g}{2}.$ Just  before landing on the platform, the monkey loses grip on the cord. After landing on the platform the monkey stays on it. The maximum extension in the elastic cord, is  $\frac{L}{x}(2+\sqrt{11+x}).\text{\hspace{0.17em}Find\hspace{0.17em}}x$

Monkey  slides down with a constant acceleration of $\frac{g}{2}.$  It means the tension in the elastic cord just before the monkey lands on the platform is  $\frac{mg}{2}$
Extension in the cord by this time $=x_0$
 Then  $k{x}_0=\frac{mg}{2}$
$\frac{4mg}{L}.x_0=\frac{mg}{2}\Rightarrow x_0=\frac{L}{8}$  
Elastic potential energy stored in the  cord $U_0=\frac{1}{2}k{\left(x_0\right)}^2=\frac{1}{2}\frac{4mg}{L}{\left(\frac{L}{8}\right)}^2=\frac{mgL}{32}$
  
 Kinetic Energy of monkey just at the time of hitting the platform $K.E=\frac{1}{2}m{V}^2=\frac{1}{2}m[2.\frac{g}{2}.\frac{9L}{8}]$
  $[\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}monkey\hspace{0.17em}\hspace{0.17em}accelerates\hspace{0.17em}\hspace{0.17em}over\hspace{0.17em}\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}length\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}L+}\frac{\text{L}}{\text{8}}\text{=}\frac{\text{9L}}{\text{8}}]$
  $\therefore K.E=\frac{9}{16}mgL$ 
 Let the cord stretch further by x
 Applying  energy conservation 
 Loss in $KE+\text{\hspace{0.17em}loss\hspace{0.17em}in\hspace{0.17em}gravitational\hspace{0.17em}\hspace{0.17em}PE=\hspace{0.17em}gain\hspace{0.17em}in\hspace{0.17em}spring\hspace{0.17em}PE}$

$$\begin{gathered} K.E+mgx=\frac{1}{2}k{(x+x_0)}^2-U_0 \\ \frac{9}{16}mgL+mgx=\frac{1}{2}k(x^2+x_0^2+2x{x}_0)-U_0[U_0=\frac{1}{2}k{x}_0^2] \\ \therefore \frac{9}{16}mgL+mgx=\frac{4mg}{2L}(x^2+2x_0.x) \\ \therefore \frac{9}{16}L+x=\frac{2}{L}{x}^2+\frac{4}{L}\frac{L}{8}.x [\therefore x_0=\frac{L}{8}] \end{gathered}$$

$$\begin{gathered} \frac{9L+16x}{16}=\frac{4x^2+xL}{2L} \Rightarrow 9L^2+16Lx=32x^2+8L.x \\ \Rightarrow 32x^2-8Lx-9L^2=0 \end{gathered}$$

$\therefore x=\frac{8L\pm \sqrt{64L^2+4\times 32\times 9L^2}}{64}$

$=\frac{L}{8}+\frac{\sqrt{19}}{8}L$  [-ve  sign is unacceptable]

$\therefore x=\frac{L}{8}(1+\sqrt{19})$

$\therefore$  Maximum extension $=x+x_0=\frac{L}{8}(2+\sqrt{19})$