Let $H$ be the height of the center O above the ground. Let $l$ be the length of the cord.
Velocity at the bottom (height $H-l$) is $v_1$.
Velocity at the top (height $H+l$) is $v_2$.
Using conservation of energy between top and bottom: $$ \frac{1}{2}m v_1^2 = \frac{1}{2}m v_2^2 + mg(2l) $$ $$ v_1^2 = v_2^2 + 4gl \quad \dots(1) $$

Cut at Bottom: The ball is launched horizontally with speed $v_1$ from height $h_1 = H-l$.
The horizontal range $R_1$ (distance from vertical axis) is: $$ R_1 = v_1 \sqrt{\frac{2h_1}{g}} = v_1 \sqrt{\frac{2(H-l)}{g}} $$ Cut at Top: The ball is launched horizontally with speed $v_2$ from height $h_2 = H+l$.
The horizontal range $R_2$ is: $$ R_2 = v_2 \sqrt{\frac{2h_2}{g}} = v_2 \sqrt{\frac{2(H+l)}{g}} $$

The problem states that points A and B are equidistant from the center O. Since A and B are on the ground, this implies their horizontal distances from the vertical axis through O are equal. $$ R_1 = R_2 $$ Substituting the range expressions: $$ v_1^2 \left( \frac{2(H-l)}{g} \right) = v_2^2 \left( \frac{2(H+l)}{g} \right) $$ $$ v_1^2 (H-l) = v_2^2 (H+l) $$ Substitute $v_1^2$ from equation (1): $$ (v_2^2 + 4gl)(H-l) = v_2^2 (H+l) $$ $$ v_2^2(H-l) + 4gl(H-l) = v_2^2(H+l) $$ $$ 4gl(H-l) = v_2^2 [ (H+l) – (H-l) ] $$ $$ 4gl(H-l) = v_2^2 (2l) $$ $$ v_2^2 = 2g(H-l) $$

The total distance between points A and B is $d$. Since they are equidistant from the center line, $R_2 = d/2$. Using the range equation for the top cut: $$ R_2^2 = v_2^2 \left( \frac{2(H+l)}{g} \right) $$ Substitute $v_2^2 = 2g(H-l)$ and $R_2 = d/2$: $$ \left(\frac{d}{2}\right)^2 = [2g(H-l)] \frac{2(H+l)}{g} $$ $$ \frac{d^2}{4} = 4(H-l)(H+l) $$ $$ \frac{d^2}{16} = H^2 – l^2 $$ $$ H^2 = l^2 + \frac{d^2}{16} $$ $$ H = \sqrt{l^2 + \frac{d^2}{16}} $$

Given values:
$l = 0.50 \text{ m}$
$d = 4.8 \text{ m}$

$$ H = \sqrt{(0.5)^2 + \frac{(4.8)^2}{16}} $$ $$ H = \sqrt{0.25 + \frac{23.04}{16}} $$ $$ H = \sqrt{0.25 + 1.44} $$ $$ H = \sqrt{1.69} $$ $$ H = 1.3 \text{ m} $$