1. The problem statement, all variables and given/known data A cannon shot a ball at 90° and reached 2.19m high. From this I need to find Vi of that and find time, Vx, Vy, Dx, and Dy of a projectile being shot from this cannon at 35.00°. I would just like to confirm I have done this correctly. Dy=2.19m θ=90.00° 2. Relevant equations Vf2-Vi2/2g Vx=Vsinθ Vy=Vcosθ Tt=2(Vsinθ)/g Vy=Vi+at Dx=Vxt Dy=Vit+at2/2 3. The attempt at a solution 2.19= 02-vi2/-19.6 Vi=6.552m/s Tt=2(6.552sin35)/9.8 Tt= 0.7670s Vx=3.758m/s Vy= 5.367m/s Dx= Vxt Dx= 3.758(0.7670) Dx=2.882m To find height at 1/2 time: d=Vit+at2/2 d=3.758(0.3835)+(-9.8x0.38352)/2 d=0.7205 I concluded with: Tt= 0.7670s Max height= 0.7205m Range= 2.882m Is this correct?