1. The problem statement, all variables and given/known data The velocity–time graph for the vertical component of the velocity of an object thrown upward from the ground which reaches the roof of a building and returns to the ground is shown in below. Calculate the height of the building. http://img832.imageshack.us/img832/9528/roofr.jpg [Broken] 2. Relevant equations s=vt/2 3. The attempt at a solution What I am thinking is that the height of the roof must be at the highest reach of the object ie. when velocity is zero. Then the total must be the total area of positive side of velocity-time graph. 1. The problem statement, all variables and given/known data The book answer is h =1/2 × 3 × 30 −1/2 × 1 × 10 = 40m Using above solution, it means the object goes higher than the roof.