   Barabolic motion angle at X time
  Hello everyoneI'm kind of new to this forum.Have been reading but not posting, so I appologise if I'm posting and this already exists. I did search about it but did not find.What I am wondering is if someone could help my with a calculation.If I have a projectile with an inicial velocity of, lets say 45m/s at an angle of 50 degrees. After 1 second of air time, what is its angle?I do know how to calculate x & y position, velocity and other relevant data but just not this.Any help would be appriciated   
  If you know the x, y position then you can calculate the angle as: arctan((y-oy) / (x-ox))where ox and oy are the position of the origin (the starting point of the projectile).   ;Quote:Original post by dmatterIf you know the x, y position then you can calculate the angle as: arctan((y-oy) / (x-ox))where ox and oy are the position of the origin (the starting point of the projectile).But wont your math give me the angle between the original position and the one after 1 sec?What I need is in which direction the projectile is traveling after 1 sec.So if it has reach its max height and is on the way down, at what angle is it doing so at the time?   ;  If you want your projectile to match the orientation of the velocity, then simply take the velocity you have already calculated and convert to an angle usingangle = atan2(velocity.y, velocity.x);  Cheers, works perfectly!   