[SOLVED] Conservation of Energy problem Assuming there is no air friction, A 1.4 kg ball is dropped from a height of 2.68 m. It hits the ground, losing 1/12 of its energy in the collision. How high will the ball bounce upward before it comes momentarily to rest? E = K sys + U sys Ef = Ei 1/2mv^2 + mgh = 1/2mv^2 + mgh I've attempted this problem but I seem to be missing some piece of information and I have run out of ideas hopefully someone can point me in the right direction. Since there are no non-conservative forces then W non conservative = 0 and there are no external forces so that Work is equal to zero. change in mechanical E = Ef = Ei Ef = Ei 1/2 m vf^2 + mghf = 1/2 m vi^2 + mghi I assumed that the final hieght is zero and the initial height is 2.68m. I also assumed that the initial velocity is 0. 1/2 mvf^2 + mg0 = 1/2 m 0 + mghi 1/2 mvf^2 = 1/2 mghi 1/2 (1.4) vf^2 = (1.4)(9.81)(2.68) i found vf^2 using kinematic equations = 52.5816 m^2/s^2 36.80712 = 36.80712 Since the final energy is 36.80712 and when it collides with the ground the energy disspates by 1/12. So : 36.80712/12 = 3.06726 So I made the the inital energy of the first bounce. Ei = Ef 3.06726 = 1/2mv^f + mgh 3.06726 = 0 + mgh 3.06726 = mgh 3.067/mg = h .22333 = h This is what I calculated the height to be but this is wrong. I'm not sure what I'm missing or if I should be looking at it in a different way.