1. The problem statement, all variables and given/known data Sam (5) kg takes off up a 50-m high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 200 N. He keeps his skis titled at 10 degrees after becoming airborne, as shown in the Figure. How far does Sam land from the base of the cliff? I drew the picture on paint. here is the link for it: http://i196.photobucket.com/albums/aa59/aliatehreem/Sam.jpg 2. Relevant equations F= ma vf^ 2 = vi ^2 + 2ad vx= v* cos theta vy= v* sin theta yf= yo + vt+ 0.5 *a* t^2 3. The attempt at a solution Find length of slope: h= 50/ (sin 10) First I have the acceleration parallel to the motion. F= ma ma= F (thrust) - weight parallel ma= 200 - mg (sine theta) a= 0.96318 Find velocity at the end of the slope. vf^ 2 = vi ^2 + 2ad vf^ 2 = 0 + 2(.96318)(50/ (sin 10) ) vf = 23.55 m/s This velocity is parallel to the slope. To find velocity after Sam gets off the slope, in the vertical and horizontal direction, the x and y component velocities are: vx= v* cos theta= 23.55* cos10= 23.2 vy= v* sin theta= 23.55* sin 10= 4.09 Since vertical displacement is -50 m, find time, and sub into horizontal component. yf= yo + vt+ 0.5 *a* t^2 -50= 0 + 4.09 t + 0.5 * -9.81 * t^2 t= 3.637 s Sub into horizontal velocity to find distance: xf= xo + vt xf = 23.2 (3.637) = 84.4 m The correct answer is 110 m. Can someone please tell me what I am doing wrong? Thanks in advance!