Given data,
[tex]\begin{gathered} m_1=2kg \\ m_2=1.34kg \\ u_1=2.28\text{ m/s} \\ u_2=0\text{ m/s (due to stationary)} \end{gathered}[/tex]From the conservation of momentum.
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Given the impact, the kickball moves forward at a rate of 1.5 m/s.
[tex]\begin{gathered} 2kg\times2.28\text{ m/s +1.34 }kg\times0\text{ m/s = 2 }kg\times v_1+1.34kg\times1.5_{} \\ 4.56+0=2v_1+2.01 \\ 2v_1=2.55 \\ v_1=1.275\text{ m/s} \end{gathered}[/tex]Thus, the velocity of the basketball after the collision is 1.275 m/s.
