1. The problem statement, all variables and given/known data A baseball is thrown from the roof of a 22.0-m-tall building with an inital velocty of magnitude 12.0 m/s and directed at an angle of 53.1 degrees above the horizontal. a) What is the speed o fthe ball just before it strikes the ground? Use energy conservation and ignore air resistance. b) What is the answer for part (a) if the inital velocity is at an angle of 53.1 below the horizontal? c) If the effects of air resistance are included, will part (a) or part (b) give the higher speed? Vo = 12 m/s y1 = 22m y2 = 0m a = -9.8 m/s² Voy=(12m/s)sin53.1 = 9.6 m/s don't know why they don't give any variables for mass of the baseball.. ;\ other problems use .145kg for baseball so i decided just to use that 2. Relevant equations E= K + Ugrav = constant K= 1/2mv1² 1/2mv1² + mgy1 = 1/2mv2² + mgy2 3. The attempt at a solution Initially, I'm guessing that the baseball only has potential energy, and when it hits the ground it's all transformed into kinetic energy. K1 + Ugrav1 = K2 + Ugrav2 0 + Ugrav1 = K2 + 0 Ugrav1 = K2 (initial gravitational potential engergy = final kinetic energy) K2 = 1/2mv2² = (1/2)(.145kg)(9.6m/s)² = 6.7J Ugrav1 = mgy1 = (.145kg)(9.8m/s²)(22m) = 31.2 J and i'm stuck. how do i find the speed? for part (b) i know that the velocity will be the same as the first part, because when it returns back from throw point, it will end up having the same velocity as if it were thrown 53.1 degrees above.