Answer
$1.414\;\rm m$
Work Step by Step
To find how far the ball will travel, we need to find its speed when released, and to find its speed when released we need to find the tension in the rope.
We know that the tension is at the minimum value which allows it to complete a full circle.
At the top of the circular path, the net force exerted on the ball is given by
$$\sum F_r=T+mg=ma_r=m\dfrac{v^2}{r}$$
where $r$ here is the length $L$ of the rope.
$$T+mg= m\dfrac{v^2}{L}$$
The minimum tension required for the ball to complete the circular path at the top point while the rope is on the verge of going slack is zero Newton.
Thus,
$$0+ \color{red}{\bf\not}mg= \color{red}{\bf\not}m\dfrac{v^2}{L}$$
Solving for $v$;
$$v=\sqrt{gL}\tag 1$$
When the rope is released at the top of the trip, we have a ball that is fired horizontally at a speed of $\sqrt{gL}$ from a height of $(1.50+0.50)=2.0\;\rm m$ and is now under the free-fall acceleration.
Now we need to find the time of the ball's trip from the moment it is released horizontally to the moment it touches the ground.
So, we can use the kinetic formula of vertical displacement.
$$y=y_0+v_{0y}t+\frac{1}{2}a_yt^2$$
where $y=0$ since it is the ground, $y_0=2$ m, $v_{0y}=0$ since it is fired horizontally, and $a_y=−g=−9.8$ m/s/s since it is under the free-fall acceleration.
$$0=2+0-4.9t^2$$
So that
$$t=\sqrt{\dfrac{2}{4.9}}\;\rm s\tag 2$$
Now we know the time of its trip and we know that its horizontal velocity component is constant and is equal to the releasing velocity. Thus, the distance is given by
$$x=\overbrace{x_0}^{=0}+v_{0x}t+\frac{1}{2}\overbrace{a_x}^{=0}t^2$$\
$$x=vt$$
Plugging from (1) and (2);
$$x=\sqrt{gL}\sqrt{\dfrac{2}{4.9}}=\sqrt{9.8\times 0.5\times \dfrac{2}{4.9}}$$
$$x=\sqrt{2}\;\rm m=\color{red}{\bf 1.414}\;\rm m$$
The ball hits the ground 1.414 m away to the right from the center of the circle.
