# A problem a bout energy

1. May 1, 2010

### zafer

when a ball with 4 kg in thrown upwards at a velocity of 40m/s, at what height is the KE equal to PE?

Last edited: May 1, 2010
2. May 1, 2010

### jtyler05si

Do you have any relevant equations yet?
Also, I assume it is being thrown from a y(0)= 0m?

3. May 1, 2010

### zafer

It is like KEi+PEi=KEf+PEf and I need to find at which height they are equal

yes it is thrown from y=0m

4. May 1, 2010

### xcvxcvvc

$$\frac{1}{2}mv_1^2 = \frac{1}{2}mv_2^2 + mgh_2$$
As the equation shows, they are equal at the height that takes half of the initial kinetic energy.

5. May 1, 2010

### zafer

can you please solve it xcvxcvvc?

6. May 1, 2010

### xcvxcvvc

Solve what? You don't use the entire equation I wrote. You solve the equation I defined with my words.

7. May 1, 2010

### jtyler05si

Have you tried anything yet?

8. May 1, 2010

### zafer

yes it should be h=40 m to be equal.
Because the KEi=3200 J and it should take the half of it to be equal,which means 1600J