1. The problem statement, all variables and given/known data A stone is projected vertically upwards from the top of a tower. It is found that its speed at a distance of x meters below the point of projection is twice its speed at the same distance above that point. Show that the stone rises a distance 5x/3metres above the top of the tower. 2. Relevant equations v2 = u2 + 2as 3. The attempt at a solution Let the velocity from the projection point upwards be v1 Let the velocity from the projection point downwards be v2 v1 = u2 - 19.6x v2 = u2 + 19.6x Since V1 = 2v2 u2 - 19.6x = 2u2 + 39.2x -58.8x = u2 I am stumped.