# Simple conservation of energy problem

1. Sep 27, 2008

### ehilge

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Conservation of energy: KEi=GPEf
.5mv2=mgh

3. 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!

2. Sep 27, 2008

### tiny-tim

Welcome to PF!

Hi ehilge! Welcome to PF!
.5(v)2=gh

Too clever by half!!

3. Sep 27, 2008

### Staff: Mentor

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.

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

4. Sep 27, 2008

### 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!