Answer:2 m/s opposite to initial direction
Explanation:
Given
mass of boy [tex]m=49.6 kg[/tex]
mass of skateboard [tex]m_2=2 kg[/tex]
speed of skateboard [tex]u=10.4 m/s[/tex]
speed of com of boy [tex]v_0=10.9 m/s[/tex]
Let v be the speed of skateboard after fall
conserving momentum
[tex](m+m_2)u=m\times v_0+m_2\times v[/tex]
[tex](49.6+2)\cdot 10.4=49.6\times 10.9+2\times v[/tex]
[tex]536.64=540.64+2v[/tex]
[tex]2v=536.64-540.64[/tex]
[tex]2v=-4[/tex]
[tex]v=-2 m/s[/tex]
v=2 m/s opposite to original direction