1. The problem statement, all variables and given/known data A diver leaves a 3m board on a trajectory that takes her 3.0m above the board, and then into the water a horizontal distance of 3.0m from the end of the board. At what speed and angle did she leave the board? 2. Relevant equations Vf^2=Vi^2 + 2ad Y=Y0 +VyT - 0.5gt^2 x=V0Cos(theta)T Y=V0Sin(theta)T - 4.9t^2 3. The attempt at a solution Alright, so I was able to get the correct angle, but I got the speed wrong. I got the angle by first applying Vf^2= Vi^2 + 2ad, I found Vi=7.667 m/s, plugged it into Y=Y0 + VyT -4.9t^2, got a quadratic equation, got the positive time of t= 1.889seconds, plugged this time into x=V0Cos(theta)T, and i got the angle of 78.04, this was correct. But how Do i find the speed from the take off? Can someone help, with an explanation and an equation please?