Given data:
* The mass of ping pong ball is 2.5 g.
* The mass of baseball is 145 g.
* The velocity of the baseball is 39 m/s.
Solution:
As the kinetic energy of the ping pong ball is same as the kinetic energy of the baseball. Thus,
![\begin{gathered} (1)/(2)m_1v^2_1=(1)/(2)m_2v^2_2 \\ m_1v^2_1=m_2v^2_2 \\ v^2_1=(m_2v^2_2)/(m_1) \\ v_1=\sqrt[]{(m_2v^2_2)/(m_1)} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/dfnosqozv26f2q5j5bwplo6dnenqngse36.png)
where m_1 is the mass of ping pong ball, m_2 is the mass of baseball, v_1 is the velocity of the ping pong ball, and v_2 is the velocity of the baseball,
![\begin{gathered} v_1=\sqrt[]{(145*(39)^2)/(2.5)} \\ v_1=297.02ms^(-1) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/akyfb4ftf8v6atmlclha2li7brco4jto9v.png)
Thus, the velocity of the ping pong ball should be 297.02 m/s.