How about negative kinetic energy?

If you get v^2 negative, then that means that the thing can't go that high.

The object will only rise to the point where KE=0, then it will go back down.

Your specific example doesn't have any units, so I'm going to have to assume them:

u = 3m/s

h = 5m

m = 2kg

If you start at 3m/s, how high will it go?

v^2 = 2 g x

x = (1/2) v^2 / g

x = (1/2) (9m^2/s^2) / 9.81m/s^2

x = 0.45m

So you can't get a final velocity for throwing a ball that slowly 5m in the air, because it will only rise about half a meter.

The answer you get (negative KE) represents how much extra energy would be required to reach that height.

This is a very confusing question. Units would be nice too.

What does u represent? I'm thinking you are describing a problem in which an object of mass 2 kg is thrown upwards with an inititial velocity, u = 3 m/s. Then you are using the energy equation to compute the velocity, v, as a function of the height, h, to which the object has risen. If you are getting negative value for v^2, that suggests you are trying to solve for the velocity at a height which is greater than it can get to with the initial velocity given.

Try solving the the height at which v = 0.

From your equation v^2=u^2 -2gh with v = 0, you get 0 = u^2 -2gh. Or 2gh = u^2. So h = u^2/(2g).

That is the maximum height the object will go up given the initial speed of u. That's about 0.5 m. So if you put in 5 m, you get a nonsensical answer.

See here's as perfect example where the math is OK, but the physics isn't.

When v^2=u^2 -2gh < 0, that's a physical impossibility. You can see this when you convert the equation back to where it came from, the conservation of energy law. In which case:

pe + ke = KE = TE; where TE is the system's total energy, which ended up with kinetic energy as the total energy after it consisted in part of potential and kinetic energies. And clearly KE = 1/2 mu^2 > 1/2 mv^2 = ke because KE = TE and ke < KE is only some fraction of TE.

So, yes, you can put in a value H > h and cause the pe to rise to PE > pe, but then that would add to the total energy that's already the max. In other words, when raising the potential energy you've exceeded the system's total energy you did a no-no, you created energy.

No negative kinetic energy as mass is positive and velocity squared is positive.

Think about this: if someone throws a rock at you its going to hurt regardless of whether the rock is thrown up at you or down at you, as it has positive kinetic energy in either case (no offense with the example, couldn't think of anything better)

When you start putting in potential energy though, potential energy is generally relative to some reference point, for example we could say zero potential energy at sea level, so you'd end up with negative energy if you went below that. I think thats how you're ending up with negative energy, its something to do with how your question is set up.