# Two-Dimensional Motion Problem

1. Jun 7, 2015

### scharry03

1. The problem statement, all variables and given/known data
A skier leaves the ramp of a ski jump with a velocity of 10.0 m/s, 15.0° above the horizontal, as shown in Figure P3.57. The slope is inclined at 50.0°, and air resistance is negligible. Find the distance from the ramp to where the jumper lands .

2. Relevant equations
tan50degrees=Yf/Xf
Yf = Yi + Vyi(t) + .5(ay)t2
Xf=Xi+Vx(t)

3. The attempt at a solution
Plugging in numbers into the first second equation using the first equation and solving for distances gave me the following: Xf(tan50) = 2.59t + -4.9t2 . Then I solved for the third equation and resulted in Xf=9.66t. Plugging this into the partially solved second equation yielded: 9.66tan(50)t = 2.59t - 4.9t2 which simplifies to -4.9t2 - 8.91t. t is supposed to equal 2.88 seconds but my solution doesn't yield that at all. What am I doing wrong? Thanks.

2. Jun 7, 2015

### scharry03

got it... the Yf i used was not negative.