Answer:
[tex]2.4375\ \text{m/s}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of cannon ball = 39 kg
[tex]m_2[/tex] = Mass of performer = 65 kg
[tex]u_1[/tex] = Initial horizontal component of cannon ball's velocity = 6.5 m/s
[tex]u_2[/tex] = Initial horizontal component of performer's velocity = 0
v = Velocity of combined mass
As the momentum of the system is conserved we have
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\dfrac{39\times 6.5+0}{39+65}\\\Rightarrow v=2.4375\ \text{m/s}[/tex]
The speed of the performer immediately after catching the cannon ball is [tex]2.4375\ \text{m/s}[/tex].
