1. The problem statement, all variables and given/known data A cannonball is launched at 90.0m/s at an angle of 60.0° above the horizontal from the top of a tower that is 38.0m high. How far away from the tower does it land if the ground is level? 2. Relevant equations Vix=90.0m/s X Cos60 Viy=90.0m/x X Sin60 3. The attempt at a solution I used the formula to find the distance from the tower, but i did not include the initial 38.0m that it was launched from. I do not know how this fits into the equation. x-comp: Vix=90.0m/s X Cos60= 45m/s y-comp: Viy=90.0m/s X Sin60=77.94228634m/s Δdy=Viy×Δt+1/2a(t)^2 0=77.94228634m/s(Δt)+1/2(-9.81m/s^2)(Δt^2) Δt=(77.94228634m/s)/(4.905m/s^2) t=15.89037438s Once you know time, you plug that in to find the horizontal distance. Δdx=Vix X Δt Δdx=(45m/s)(15.89037438s) Δdx= 715.0668471m This is the answer because I did not include the 38.0m. How can i do that? The correct answer is 736m.