1. The problem statement, all variables and given/known data A baseball thrown at an angle of 60.0 above the horizontal strikes a building 18.0m away at a point 6.00m above the point from which it is thrown. Ignore air resistance. A) Find the magnitude of the initial velocity of the baseball (the velocity with which the baseball is thrown). B) Find the magnitude of the velocity of the baseball just before it strikes the building. C) Find the direction of the velocity of the baseball just before it strikes the building. known: initial angle = 60o x = 18, y = 6 @ point of impact a = 9.8 m/s2 2. Relevant equations x = v0x*t y = v0y*t + .5*a*t2 3. The attempt at a solution I began by manipulating the equations above to create an equation that solves for the magnitude of the velocity, v at x=18, y=6 : 18 = cos(60)*v*t 6 = sin(60)*v*t + .5*a*t2 t = 18/(cos(60)*v) 6 = sin(60)*v*(18/(cos(60)*v))+.5*-9.8*(18/(cos(60)*v))2 v = sqrt((.5*-9.8*182)/(6-(18*sin(60)/cos(60))))/cos(60) And using this I found v to be 15.9, which is incorrect. I don't think the rest will be any problem if I can figure out v, but I can't find what I'm doing wrong for the life of me. Also, if there is a different approach I should be taking please tell me, thanks!