I need help on projectile motion questions because i had to miss a whole lesson for some reason and my teacher said to try some of the questions. Anyways im stuck on this question. A cannon is set at an angle of 45 degrees above the horizontal. A cannon ball leaves the muzzle with a speed of 220 m/s. Air resistance is negligible. Determine the cannonball's a) maximum height b) time of flight c) horizontal range (to the same vertical level) This is what i did. Im not going to do step by step on the part i got. Before these i started i found VIx which is 155.6 m/s and Viy is the same a) At max height Vfy=0( i dont get why but i know thats how you do it) 2ad=Vi^2 + Vf^2 d= (155.6)^2/-2(-9.8) d= 1240m b)i have no idea how to get this one i hope you can help me. c)i started but i couldnt finish. This is what i did d=Vit + 1/2a(t^2) 1240=155.6t - 4.9t^2 4.9t^2 - 155.6t + 1240 = 0 I know i need to do the quadratic formula after this but i get a negative with the radical so i dont know what to do i think i made a mistake somewhere. Can you please help me?