## A problem a bout energy

when a ball with 4 kg in thrown upwards at a velocity of 40m/s, at what height is the KE equal to PE?
 Do you have any relevant equations yet? Also, I assume it is being thrown from a y(0)= 0m?
 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

## A problem a bout energy

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

 Quote by zafer can you please solve it xcvxcvvc?
Solve what? You don't use the entire equation I wrote. You solve the equation I defined with my words.
 Have you tried anything yet?
 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