Finding height where projectile has half initial speed

[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.

 

[Parth Dave]

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

 

[senseandsanity]

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

 

[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?

 

[Doc Al]

Using conservation of energy, I found the maximum height v to which an object will rise is h_max= ((1/2)*(v^2))/(g).

I assume you mean that you found the maxiumum height for an object projected straight up at initial speed v. If so, good.

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.

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.