physics daredevil, wooo!! 1. The problem statement, all variables and given/known data A Hollywood daredevil plans to jump the canyon shown here http://img39.imageshack.us/img39/9406/jonesf336.gif [Broken] on a motorcycle. If he desires a 3.0-second flight time, what is a) the correct angle for his launch ramp, b) his correct launch speed, c) the correct angle for his landing ramp, and d) his predicted landing speed? (Neglect air resistance) Also, in the picture, the ramp length is 60m. I'm not sure if that makes a difference. 2. Relevant equations yf=y0+voyt-1/2 gt^2 vf^2=voy^2-2gy vy=voysintheta 3. The attempt at a solution so since this flight is all due to velocity in the y direction, we shouldn't consider vx because it's zero all the way, right? so we're left with voy. we know that yo is 15m, yf is 0m, and t is 3.0s so 0=15+voy(3)-4.9(9) voy=9.7m/s so then you plug that into equation #2: I'm not completely sure if in this case y=15m vf^2=9.7^2-2(9.8)(15) but then I get vf^2=-199.91 and that's not possible. vf=14.14m/s? and if vf really is 14.14m/s, how do I find the angle?