I must be missing something because I keep finding that this cannon ball falls short. Anyway, the question is at what angle does a cannon have to fire for the cannon ball to hit the ground 269 ft away if the initial velocity of the ball is 100ft/s? I set up 2 equations like so.
100tcos(x)=269
100sin(x)-32.2t=0
t is the time and x is the angle, someone let me know that my brain just isn’t working today
Nevermind, it should be t/2 in the second equation, I’m stupid,nothing to see here.
Apparently my brain wasn’t working properly either becuase it took me a while to get the right answer. Your mistake is that your 1st equation is solving for the final position of the ball while your 2nd is solving for a point of 0 velocity i.e. the apex of the trajectory. Instead of 269 in your first equation put 269/2 and you should get the right answer.
t=1.55s x=.52375 radians or 30 degrees.
Bah I see you got it on preview damn me for taking too long. I am posting it anyhow.
I think the parametric equations for the vacuum trajectory are:
x = v[sub]o[/sub]tcos(a)
y = v[sub]o[/sub]tsin(a) - gt[sup]2[/sup]/2
x is horizontal distance, y is vertical distance, v[sub]o[/sub] is initial velocity and a is elevation angle.
Oh yes, g is the acceleration of gravity.
The second equations says that in the absence of gravity the vertical distance of the ball will be along the straight line v[sub]o[/sub]tsin(a). Gravity makes it fall below that distance by gt[sup]2[/sup]/2.
Use more gunpowder, that should help.
I took a little detour to the hospital so I’m a little late with this.
The problem has two solutions:
30 deg, time of flight 3.11 sec.
53.89 deg, time of flight 5.02 sec.
For any range less than the maximum, which comes at 45 deg. elevations, there will always be two solutions. One at more and one at less than 45 deg.
Max range is about 311 ft.