1. The problem statement, all variables and given/known data A ball is launched from a 35.0m high cliff with an initial velocity of 47.0m/s at an angle of 39.0 degrees above the horizontal. On the way down it just clears a 2m high wall. A) How far away is the wall? 2. Relevant equations Delta D=Vx*T + 1/2*a*t^2 3. The attempt at a solution I calculated the Verticle velocity to be 36.53m/s and the horizontal velocity to be 29.58m/s. Using the Velocity in the in the y direction, and the acceleration due to gravity, I used this equation to calculate time; Df-Di=Vy*t + 1/2a*t^2 4.905m/s^2*T^2+36.53m/s*t-33.0m Then used the quadratic equation to get 0.813985s I plugged this into D=Vx*T and got 24.07m I figured I mixed up my signs somewhere along the long so I re did it and got 8.26s and then got 244.4m. The correct answer is 256m.. I'm really close but I don't know why I'm 12 off.. I checked everything over. Thanks!!