# Conservation of energy and maximum height

1. Nov 23, 2004

### senseandsanity

Using conservation of energy, I found the maximum height v to which an object will rise is h_max= ((1/2)*(v^2))/(g).
At what height h above the ground does the projectile have a speed of 0.5v?
I found h= (0.125*(v^2))/(g) but that isn't correct.

2. Nov 23, 2004

### Parth Dave

I think you did it right, but made a small algebraic blunder. Check over your work clearly.

3. Nov 25, 2004

### senseandsanity

I'm still not getting the right answer and I need it for another question. Any other help would be great.

4. Nov 25, 2004

### Sirus

Can you enlighten us a little on the details of the problem? What exactly does v represent in these equations? How do you get them?

5. Nov 26, 2004

### Staff: Mentor

I assume you mean that you found the maxiumum height for an object projected straight up at initial speed v. If so, good.
What fraction of the original KE does the object have when its speed is 0.5v? So how much of its final PE does it have? Use that to figure the height in terms of h_max.