Last problem of the year!

Just before the particle collides with the ground, it will have a speed $\sqrt{2gH}$ and the string will make an angle

$\theta = \arcsin(\frac{H}{L})$

Since the ground is inelastic and horizontal, the vertical component of that speed will be lost, and the particle will then be traveling along the ground with a speed of

$V_{ground}=\sqrt{2gH}\sin(\theta)=\frac{H\sqrt{2gH}}{L}$

The particle will travel along the ground to the other side of A and eventually the string will again make an angle $\theta$ with the ground. Since the string is presumably inextensible, the particle will lose all of it's speed in the direction parallel to the string. Therefore immediately after the string becomes taut and the particle begins to leave the ground, it will have a speed of

$V_{ground}\sin(\theta)=\left(\frac{H}{L}\right)^2\sqrt{2gH}$

From now on, the string only applies a force perpendicular to the particle's motion, so energy is conserved and we can write the equation

$\frac{m}{2}\left(\left(\frac{H}{L}\right)^2\sqrt{2gH}\right)^2=mgx$

Which means

$x=\frac{H^5}{L^4}$

Bro please change the choices,, one can simply use dimensional analsys to solve this,