1. The problem statement, all variables and given/known data A man stands on the roof of a building that is 30.0 m tall and throws a rock with a velocity of magnitude 60.0 m/s at an angle 33 degrees from the horizontal. calculate (a) the maximum height above the roof reached by the rock. (b) the magnitude of the velocity of the rock just before it strikes the ground. here's the figure :D http://img245.imageshack.us/img245/7768/doc1.png [Broken] 2. Relevant equations V^2 = Vo^2 + 2gd V= Vo + gt Tan (teta) = Vx/Vy We use 10 for gravity, not 9.8. 3. The attempt at a solution Vox = 60cos33 = 50.32 m/s Voy = 60sin33 = 32.67 m/s (a) V^2 = Vo^2 + 2gd 0 = (32.67)^2 + 2(-10) d hmax = 53.39 m I got maximum height (hmax) correct :D but I'm having a hard time with magnitude of the velocity T T total time of travel to hmax V=Voy + gt tmax= 3.268 s total time of travel tmax x 2 t = 6.54 s since: tan (teta) = Vy/Vx Vx = Vox = 50.32 m/s Vy = ? Vy = Voy - gt = 32.67 - 10 (6.54) Vy = -32.73 m/s tan (teta) = Vy/Vx teta = tan^-1 | Vy |/| Vx | = tan^-1 |-32.73 m/s| / | 50.32 m/s | teta= 33.04 degrees V = SQUARE OF (Vx^2 + Vy^2) = SQUARE OF ( 50.32^2 + -32.73^2) = SQUARE OF ( 3,603.3553 V = 60.027 m/s therefore : 60.027 m/s, 33.04 degrees, 2nd quadrant. is this the correct answer? :D this is what I solved but I'm not confident if it's really correct :) please can someone help ?