Answer:
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
[tex]V1 = \frac{ m*V(m, g) + M*Vo}{M+m}[/tex]
On substituting the values we get
V1 = [tex]\frac{80*5.6 + 37.5*0}{80+ 37.5}[/tex]
V1 = [tex]\frac{448}{117.5}[/tex] = 3.812 m/s
Therefore the velocity of the boat and fisherman after he lands in the boat is 3.812m/s.
