An 80.0 kg fisherman jumps into a 37.5kg boat that is at rest. If the velocity of the fisherman is 5.60m/s when he jumps from the dock, what will be the velocity of the boat and fisherman after he lands in the boat?

The velocity of the boat and fisherman after he lands in the boat is 3.812m/s

Explanation:

We will solve this question on the general Derivation for general case of a man jumping n the boat

m = Mass of the man = 80kg

M= mass of the boat 37.5kg

V(m, g) = Velocity of man with respect to ground = 5.6m/s

V(b , g) = Velocity of boat with respect to ground = Vo = 0m/s ( rest)

velocity of the boat and fisherman after he lands in the boat = V1

As there is no external force acting on the man then we can apply conservation of the Linear Momentum.

m*V(m, g) + M*Vo = (M+m)V1

 On substituting the values we get

V1 =

V1    = = 3.812 m/s

Therefore the velocity of the boat and fisherman after he lands in the boat is 3.812m/s.