Well we could assume that the force up from the slush on the ball is constant and that the kinetic energy of the ball goes into work, force times depth
F d = (1/2) m v^2 = m g h
if d is twice then h is twice
if drop down an stone -that it is like a ball- into the slush and if it was soft enough, stone come into the slush in some distance, if throw it away from a certain height to comes into slush and if we want stone to come into slush 2 times more than before, from what hight should we throw it away?
1 answer