Aboy of 50 kg jumps off a boat (30 kg), causing the boat to move to the right at 2.0 m/s. in what direction and at what velocity does the boy move?

For this problem, we will assume the following:
1- The boy an the boat were originally at rest (V=0 m/sec)
2- The right direction is our default direction having the positive sign while the left direction will be having a negative sign

Now, the law of conservation of momentum states that: "For a closed system, momentum must be conserved"
This means that:
M1V1 + M2V2 = M3V3
where:
M1 is the mass of boy = 50 kg
V1 is the velocity of the boy we want to find
M2 is the mass of boat = 30 kg
V2 is the velocity of boat = 2 m/sec
M3 is the mass of boy and boat together = 30+50 = 80 kg
V3 is the velocity of boy and boat together = 0 m/sec

Substitute with the givens in the above equation to get V1 as follows:
50V1 + 30(2) = 80(0)
50 V1 + 60 = 0
50 V1 = -60
V1 = -1.2 m/sec

This means that:
velocity of boy is 1.2 m/sec in the left direction

Hope this helps :)

Iassume the answer is 75 feet. inwards to outwards rather than outwards to inwards. probably be 75.hachures are stokes drawn in direction of the steepest slope. a series of short parallel lines drawn on a map to indicate topographic relief. it is actually called hachures on contour lines. the small marks in the center of the feature are called hachures.observe the picture attached