1. A 0.43 kg rock is projected from the edge of the top of a building with an initial velocity of 9.2 m/s at an angle 45◦ above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 14.9 m from the base of the building. How tall is the building? Assume the ground is level and that the side of the building is vertical. The acceleration of gravity is 9.8 m/s2 . 2. Relevant equations 3. The attempt at a solution I've broken the problem down into halves. I've solved the top half ( the part of the motion equal in height to the roof ) but don't know how to get the second half ( the part of the motion after the rock is past the point level with the roof. ) Have I even gone about this right? First half: Vxi=9.2cos45=6.51m/s Vyi=9.2sin45=6.51m/s delta t=(Vfy-Viy)/g =1.33 s delta d=Vix(delta t) =8.65 m Second half: (Not sure if right) Vix=6.51cos45=4.6m/s Viy=6.51sin45=4.6m/s Now I am stuck!