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.
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:
I solve for time on the horizontal time in terms of x.
I can now use that t and plug it into the vertical t, ensuring that x is the only variable and solve for x.
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!!