How Does Batman's Jump Impact the Speed of a Getaway Boat?

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wallace13
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Batman (mass = 105 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +15 m/s. What is the velocity of the boat after Batman lands in it?
.5mv squared + mghf = .5mVi squared + Mghi.5 x 635 x Vf squared = .5 x 530 x 15 squared

= 13.7m/s
 
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I mean that's quite a funny and not specific problem. Does batman jump vertically from the bridge? do you mean this by "straight down"? If yes then ok. What is the mass of the criminal? Or this the given "boat" the sum of the mass of the boat itself and the criminal?
You should be a bit more specific... :D :D
Now is the water considered "solid", because otherwise I could well imagine that someone with fifth the mass of a boat jump on it, and it collapsed or submerged into the water, which is another case...:D this should be specified too.. :D
(I am not "bugging" with these, just in the future it is better if you specify these, because then it will be much more comprehendible :D)

If we have all these. Then we can go onto the solution. Now why would you say that energy is conserved in this case? The "collision" between Batman and the boat is totally non-elastic (If we consider the water to be "solid"), hence energy is not conserved, so your solution is wrong.
But if we consider batman to jump verticaly then the horizontal momentum is conserved, since there is no horizontal force.

So use the conservation of momentum in the horizontal direction.
 
I copied the problem straight out of the book lol our book just sucks
 
wallace13 said:
Batman (mass = 105 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocity of the boat is initially +15 m/s. What is the velocity of the boat after Batman lands in it?

.5mv squared + mghf = .5mVi squared + Mghi

.5 x 635 x Vf squared = .5 x 530 x 15 squared

= 13.7m/s

This is not a kinetic energy problem. The potential energy from his jump does not affect the combined horizontal motion.

What's asked is for you to apply the conservation of momentum.

Mass X Velocity before = Mass X Velocity after