What height is KE equal to PE for a 4kg ball thrown at 40m/s?

[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?

 

[jtyler05si]

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

 

[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

 

[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.

 

[zafer]

can you please solve it xcvxcvvc?

 

[xcvxcvvc]

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.

 

[jtyler05si]

Have you tried anything yet?

 

[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