Answer:
![2.4375\ \text{m/s}](https://img.qammunity.org/2022/formulas/physics/college/a14j2573xze5qflrsgbhrf3tery3b5p3ii.png)
Step-by-step explanation:
= Mass of cannon ball = 39 kg
= Mass of performer = 65 kg
= Initial horizontal component of cannon ball's velocity = 6.5 m/s
= Initial horizontal component of performer's velocity = 0
v = Velocity of combined mass
As the momentum of the system is conserved we have
![m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=(m_1u_1+m_2u_2)/(m_1+m_2)\\\Rightarrow v=(39* 6.5+0)/(39+65)\\\Rightarrow v=2.4375\ \text{m/s}](https://img.qammunity.org/2022/formulas/physics/college/nojz37oag7kga67klymsu7fbf4cwn9bjb0.png)
The speed of the performer immediately after catching the cannon ball is
.