Answer:
  v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
       p₀ = m v₁₀
final. After the crash
        = m
 = m  + m v_{2f}
 + m v_{2f}
Recall that velocities are a vector so it has x and y components
        p₀ = p_{f}
we write this equation for each axis
X axis
        m v₁₀ = m v_{1fx} + m v_{2fx}
        
Y Axis  
        0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
       sin 23.3 = v_{2fy} / v_{2f}
       cos 23.3 = v_{2fx} / v_{2f}
       v_{2fy} = v_{2f} sin 23.3
       v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
        m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
        0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
       1.83 = v_{1f} cos θ + 1.15 cos 23.3
        0 = - v_{1f} sin θ + 1.15 sin 23.3
       1.83 = v_{1f} cos θ + 1.0562
         0 = - v_{1f} sin θ + 0.4549
      v_{1f} sin θ = 0.4549
      v_{1f}  cos θ = -0.7738
we divide these two equations
       tan θ = - 0.5878
       θ = tan-1 (-0.5878)
        θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
         v_{1f} cos (-30.45) = - 0.7738
         v_{1f} = -0.7738 / cos 30.45
         v_{1f} = -0.8976 m / s
the component and this speed is
        v_{1fy} = v1f sin θ
        v_{1fy} = 0.8976 sin (30.45)
        v_{1fy} = - 0.4549 m / s