Answer :
Answer:
-0.912 m/s
Explanation:
When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.
[tex] (m_c + m_b)v_b + m_pv_p = 0[/tex]
where [tex]m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg[/tex] are the mass of the child, the boat and the package, respectively. [tex], v_p = 10m/s, v_b[/tex] are the velocity of the package and the boat after throwing.
[tex] (24.6 + 39)v_b + 5.8*10 = 0[/tex]
[tex]63.6v_b + 58 = 0[/tex]
[tex]v_b = -58/63.6 = -0.912 m/s[/tex]