Recent content by benrocks1

thanks so much for your all your help and time!

Oh so the kinetic energy is 0 because it is not in motion?

I understand that mgh would = 0 when it hits the ground, but why is it 0 before the drop?

would 1/2mv^2-mgh remain the same?

the potential energy will decrease by x. does that mean that 1/2mv^2-mgh also decreases by x?

So 1/2mv^2-mgh would also increase by x?

While the kinetic energy increases, the potential energy will decrease. Is that correct? I've thought about it, but i still do not understand why I would use 1/2mv^2-mgh

I know that 1/2mv^2 is Kinetic energy and mgh is potential energy. My question is in some example problems they use 1/2mv^2 and in others they use 1/2mv^2-mgh.

Can someone explain to me when you use 1/2mv^2-mgh as opposed to 1/2mv^2 or mgh