Final answer:
The velocity of the boat immediately after the package is thrown is 10.0 m/s in the opposite direction to the package.
Step-by-step explanation:
To solve this problem, we can apply the principle of conservation of momentum. The initial momentum of the system (boat + child) is zero since both are initially at rest.
The momentum of the package can be calculated using the equation:
momentum = mass × velocity
Since the package is thrown horizontally, its velocity is only in the x-direction. Therefore, the momentum of the package is (6.6 kg) × (10.2 m/s) = 67.32 kg·m/s.
According to the conservation of momentum, the final momentum of the system (boat + child + package) must also be zero.
Let's assume the velocity of the boat after the package is thrown is v.
The momentum of the boat is given by:
momentum = mass × velocity
The child's momentum is given by:
momentum = mass × velocity
The momentum of the system after the package is thrown is the sum of the momenta of the boat, child, and package.
We can express this as an equation:
(mass of boat × v) + (mass of child × 0) + (mass of package × 10.2 m/s) = 0
Since the mass of the child is 29.9 kg and the mass of the boat is 44.7 kg, the equation becomes:
(44.7 kg × v) + (29.9 kg × 0) + (6.6 kg × 10.2 m/s) = 0
Solving this equation, we find that the velocity of the boat after the package is thrown is approximately -10.0 m/s.
The negative sign indicates that the boat is moving in the opposite direction to the package.
Therefore, the velocity of the boat immediately after the package is thrown is 10.0 m/s in the opposite direction to the package.