Simple conservation of energy problem

[ehilge]

Homework Statement

All I need to do is figure out at what height h above the ground does a projectile have .5v using only v and g in the answer.

Homework Equations

Conservation of energy: KEi=GPEf
.5mv²=mgh

The Attempt at a Solution

so I went ahead and just plugged in .5v or v and solved for h so...
.5(.5v)²=gh
.5(.25v²)=gh
.125v²=gh
h=v²/8g

however, the correct answer is 3v²/8g

so I guess all I need to know is where the 3 came from. I don't see any algebraic errors and I frankly don't have any other ideas
Any help is greatly appreciated,
Thanks!

 

[tiny-tim]

Welcome to PF!

Hi ehilge! Welcome to PF!

.5(.5v)²=gh

.5(v)²=gh

Too clever by half!

 

[Doc Al]

You need to set up the conservation of energy equation properly. What's the total energy when the projectile just leaves the ground? What's the total energy when its speed = 0.5v and its height = h? Set those equal.

Homework Equations

Conservation of energy: KEi=GPEf
.5mv²=mgh

In general, this is not true. It's only true if all the KE is transformed into PE.

 

[ehilge]

ok, so I got the correct answer when I set .5mv²=.5m(.5v)²+mgh. I think I actually just read my own question wrong. I thought I was trying to find the height if the initial velocity is .5v but it looks like I needed to find what the height is when the velocity is .5v but the intitial is still v. Oh well, I geuss I better pay more attention next time. Thanks for you help!