# Problem about conservation of energy

1. Apr 3, 2006

### physstudent1

I have a problem about conservation of energy I did it but I'm not sure if it is correct...

"From 2.15m a ball is thrown upwards with KE of 5.4 J it's mass is .6 kg will it go 3.04 m high? use the law of conservation of energy"

What I did was set KEinitial + PEinitial = KEfinal + PEfinal

eventually ending up with v = .2357 m/s now coming to that I thought that maybe that means it will reach the height because It still has speed at the designated height but then I realized its not velocity and if it was going down it would still have a positive speed...can anyone check this out for me how do I know if it will reach that height I dont' really get it...I used the designated 3.04 for the PEfinal calculation..

2. Apr 3, 2006

### KD

For this problem, a diagram would be best. Point a would be ball at 2.15m (at rest). Make this your reference leve. And then there is point b, some point higher than point a, assuming its the apex. KEa + PEa = KEb + PEb. Now, there is no potential energy at the reference level so there is only KEa left on that side. You already said that was equal to 5.4 J. On the other side, for point b, there is no kinetic energy because the ball has reached its apex. The velocity would be zero, so you would only have PEb left. That equals mgh. The mass is .6 and g is 9.81.

So you have 54J = .6X9.81h. Solve for h. That says how high the ball went. Add that number to your 2.15m. Is it greater than 3.04m?

3. Apr 3, 2006

### physstudent1

wow thanks

Wow I never thought to start the refernce line at 2.15 thats a great idea thank you very much I ended up getting 3.06 for the total height after adding it to 2.15 which means it does reach the 3.04m I think thats right? thank you very much!

4. Apr 3, 2006

### KD

Yeah, I got 3.07, along the same answer as you, so yes, it did reach 3.04m. Glad I could help.