The mechanical energy of the ball is conserved because the tension force is always perpendicular to the velocity, and gravity is a conservative force. Let the potential energy be zero ($U=0$) at the lowest point of the swing.

The initial energy $E_i$ when the ball is released from the horizontal position (height $L$ above the lowest point) is purely potential:

$$E_i = K_i + U_i = 0 + mgL = mgL$$

[A1] At the lowest point, the height is zero, and the energy $E_{low}$ is purely kinetic. Let the speed be $v_{low}$.

$$E_{low} = K_{low} + U_{low} = \frac{1}{2}mv_{low}^2 + 0$$

By conservation of energy, $E_i = E_{low}$:

$$mgL = \frac{1}{2}mv_{low}^2$$ $$v_{low} = \sqrt{2gL}$$

[A2] After the string hits the peg, the ball swings around the peg in a circle of radius $r = L-d$. The speed at the lowest point remains $v_{low}$. The highest point of this new swing is a vertical distance of $2r$ above the lowest point. Let the speed at this highest point be $v_{high}$. The energy $E_{high}$ at this point is:

$$E_{high} = K_{high} + U_{high} = \frac{1}{2}mv_{high}^2 + mg(2r)$$ $$E_{high} = \frac{1}{2}mv_{high}^2 + 2mg(L-d)$$

By conservation of energy, $E_{low} = E_{high}$:

$$\frac{1}{2}mv_{low}^2 = \frac{1}{2}mv_{high}^2 + 2mg(L-d)$$

Substituting $v_{low}^2 = 2gL$:

$$mgL = \frac{1}{2}mv_{high}^2 + 2mg(L-d)$$ $$\frac{1}{2}v_{high}^2 = gL - 2g(L-d) = gL - 2gL + 2gd = g(2d-L)$$ $$v_{high} = \sqrt{2g(2d - L)}$$

This requires $2d \ge L$ for the ball to reach the top.