Answer :
Answer:
The velocity of the boat after Batman lands in it is 6.97 m/s.
Explanation:
Given that,
Mass of batman, m = 88.6 kg
Mass of boat, m' = 484 kg
The velocity of the boat is initially +8.25 m/s, v' = 8.25 m/s
We need to find the velocity of the boat after Batman lands in it. In this case, the conservation of linear momentum works. So,
Initial momentum = final momentum
[tex]mv+m'v'=(m+m')V[/tex]
v = 0 (as the batman was at rest initially)
[tex]m'v'=(m+m')V\\\\V=\dfrac{m'v'}{(m+m')}\\\\V=\dfrac{484\times 8.25}{(88.6+484)}\\\\V=6.97\ m/s[/tex]
So, the velocity of the boat after Batman lands in it is 6.97 m/s.