I am trying to solve for Vy with given Vx. I know how to do this for the BLUE trajectory line, but I do not know how to solve this if there is an obstruction in the way (like the gray wall).
My givens are:
Do I need to solve for the blue trajectory then adjust that for the red trajectory? If so, wouldn't I be adjusting both Vx and Vy in that case? How do I do that?
It's possible this is very simple and I'm just not seeing it lol Also, I'm not concerned about the angle of projection here, I am literally trying to solve for the X and Y components of the initial Velocity. So unless it's absolutely necessary, let's try to keep Trigonometry out of this. (Me and Trig aren't friends)
ETA: I am also not concerned about time (t).
ETA: Do not assume Y2 is the max height of the trajectory, It's only the height of the wall. So X1 could be really small while X2 is really large - placing the wall really close to the projectile's initial position.
46 KB Views: 1,161