1. The problem statement, all variables and given/known data Hi, I am doing an end of the year calculus presentation on projectile motion in human cannonballs. As an introduction to the projectile motion equations, my group found this problem: The original problem with a figure of it can be found http://higheredbcs.wiley.com/legacy/college/anton/0471482730/calc_horizons/blammo.pdf" [Broken] I'm not a physics student, so thank you for any help! 2. Relevant equations I'm sorry, but I couldn't figure out the latex equations...Hope these are clear enough. Projectile Motion Formulas R=(vo2/g)sin(2q) r(t)=v0xti+(voyt-.5gt2j which, if I'm correct (??) breaks down into: x(t)=xo+voxt y(t)=yo+voyt-.5gt2 (Are there equations I'm missing for this problem that would be more useful/better than these??) 3. The attempt at a solution Okay...Here we go: R=(vo2/g)sin(2q) For the range of the cannon (R) to be maximum, the angle of elevation (q) must equal 45o. R=(352/9.798)sin(2*45) R=125.026m x(t)=xo+voxt y(t)=yo+voyt-.5gt2 @45o, vox=voy. vox=vocos(q) vox=voy=24.749m/s I placed the wall of fire at x=0. So, solving for xo would give the distance from the wall the cannon must be placed. (Correct??) 0=xo+voxt t=-xo/vox Sub t into the y(t) equation: y(t)=20=5+voyt-.5gt2 15=-xo-4.899(-xo/24.749)2 xo=-17.43m R+xo=length of pool The pool is 107.596m long. So, I think I have solved this correctly. One thing I'm worried about is the assumption of q=45o. I know this makes R the greatest, but does it make the length of the pool the greatest? If the angle was steeper, could the cannon be placed closer to the fire, making the pool be a greater part of the path? Any help or confirmation of my answers would be greatly appreciated. Thank you so much!!