Answer
$7.94\;\rm m/s$
Work Step by Step
We have here two stages, the perfect elastic collision between the two balls, and the circular motion of the second ball.
In the first stage, we can use the conservation of momentum to find the velocity of the second ball just after the collision and just after it starts to move in the circular path.
In the second stage, we can use the conservation of energy of the second ball to find its final height.
In the first stage, while the second $m_2$ was still at rest, the collision is considered perfectly elastic, so the velocity of the second ball after the collision is given by
$$v_{fx,2}=\dfrac{2m_1v_0}{m_1+m_2} $$
Thus,
$$v_0=\dfrac{m_1+m_2}{2m_1}\;v_{fx,2}\tag 1$$
In the second stage, the energy is conserved after the collision,
$$E_i=E_f$$
$$K_{i2}+U_{ig,2}=K_{f2}+U_{ig,2}$$
Note that, when the ball 2 reaches its maximum height, its speed is zero. The initial gravitational potential energy of ball 2 is zero.
$$K_{i2}+0=0+U_{ig,2}$$
$$\frac{1}{2}\color{red}{\bf\not} m_2v_{fx,2}^2 =\color{red}{\bf\not} m_2gh $$
$$ v_{fx,2}^2 =2gh\tag 2$$
where $h$, as seen in the figure below, is given by
$$\cos\theta=\dfrac{R-h}{R}$$
Thus,
$$h=R-R\cos\theta $$
Plugging into (2);
$$ v_{fx,2}^2 =2g(R-R\cos\theta )=2gR(1-\cos\theta) $$
$$v_{fx,2}=\sqrt{2gR(1-\cos\theta)} $$
Plugging into (1);
$$v_0=\dfrac{m_1+m_2}{2m_1} \sqrt{2gR(1-\cos\theta) }$$
Plugging the known;
$$v_0=\dfrac{0.02+0.1}{2\times 0.02} \sqrt{2\times 9.8\times 1(1-\cos50^\circ) }$$
$$v_0=\color{red}{\bf 7.94}\;\rm m/s$$
