1. The problem statement, all variables and given/known data If a trajectory is shot at angle theta at a fixed projectile velocity of 50 m/s, what angle(s) could it be shot so that it just makes it over a wall 30 meters high, 220 meters down the path. 2. Relevant equations d=1/2at^2+vt sinx*v=vertical component cosx*v=horizontal component 3. The attempt at a solution Time connects the horizontal and vertical distances, so my approach is to solve for time and then substitute. On the vertical end, we can say: Vo=50*sinx a=-9.8 d=30m Horizontal: Vo=50*cosx Vf=50*cosx a=0 d=220 I solve for time on the horizontal time in terms of x. d=vt 220=50cosx*t t=220/50cosx I can now use that t and plug it into the vertical t, ensuring that x is the only variable and solve for x. d=1/2at^2+Vot 30=1/2(-9.8)(220/50cosx)^2+50sinx(220/50cosx) After setting that up, I am stuck. I have tried it multiple times, combining them and using LCD, etc. I end up with a cosx and a sinx and can't use trig identities or anything to isolate x. I can't find any way to solve for the angle of launch Thanks for the help!!