Vertical height reached after $4s$ is

$h=ut -\large\frac{1}{2}$$gt^2$

$\quad=50 \times 4 -\large\frac{1}{2}$$\times 10 \times 16 =120 m$

Percentage of energy converted

$\quad=\large\frac{mgh}{\Large\frac{1}{2}mv^2} $$\times 100$

$\quad=\large\frac{2 \times 10 \times 120}{50 \times 50}$$ \times 100$

$\quad=96 \;\%$

Hence a is the correct answer.